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Nuclear reactions - class-XII

Description: nuclear reactions
Number of Questions: 103
Created by:
Tags: physics nuclei atomic nuclei
Attempted 0/103 Correct 0 Score 0

Which of the following relations is correct?

  1. $E = mc$

  2. $E = mc^2$

  3. $E = 2mc^2$

  4. $E = mc^2/4$


Correct Option: B
Explanation:

According to Einstein, energy and mass are related by the relation.
$E = mc^2$
where c is the speed of light in vacuum.

One milligram of matter is converted into energy. The energy released will be

  1. $9\times 10^{6} J$

  2. $9\times 10^{8}J$

  3. $9\times 10^{10}J$

  4. $9\times 10^{12}J$


Correct Option: C
Explanation:

Here, $m = 1\ mg = 1\times 10^{-3} g $

             $= 1\times 10^{-6}kg$
According to Einstein mass-energy equivalence
                $E = mc^{2}$
where $c$ is the speed of light in vacuum
$\therefore E = (1\times 10^{-6} kg)(3\times 10^{8}ms^{-1})^{2}$
$ = 9\times 10^{10}J$.

The relation between the volume $V$ and the mass $M$ of a nucleus is:

  1. $V\propto M^{3}$

  2. $V\propto M^{1/3}$

  3. $V\propto M$

  4. $V\propto 1/M$


Correct Option: C
Explanation:

Since the density of nucleus is fixed.

$D=\cfrac { M }{ V } \ \Rightarrow M=DV\ M\propto V$

A student wrote the relation for one unified atomic mass unit (u) as $1u=931.5MeV$. What is the correct relation?

  1. $1 u\times c=931.5 MeV$

  2. $1 u\times c^2=931.5 MeV$

  3. $\dfrac{1u}{c^2}=931.5 MeV$

  4. $(1u)^2\times c=931.5Me V$


Correct Option: B
Explanation:

$1u=931.5 MeV$
we know $1u=1.66\times 10^{-21}kv$
$931.5 MeV=931.5\times 10^6\times 1.6\times 10^{-19}J=1.49\times 10^{-10}$
$E=(1u)(c^2)=(1.66\times 10^{-27})(3\times 10^8)^2J=1.494\times 10^{-10}J$
so $(1u)(c)^2=931.5MeV$

A nucleus of mass number $A$ originally at rest emits $\alpha$- particle with speed $v$. The recoil speed of daughter nucleus is:

  1. $\cfrac{4v}{A-4}$

  2. $\cfrac{4v}{A+4}$

  3. $\cfrac{v}{A-4}$

  4. $\cfrac{v}{A+4}$


Correct Option: A

As the mass number increase, binding energy per nucleon,

  1. increases

  2. decreases

  3. remain same

  4. may increase or may decrease


Correct Option: D

Per nucleon energy of $ _ { 3 } L ^ { 7 }$ and $2 ^ { \mathrm { H } e ^ { 4 } }$  nucleus is 5. 60  MeV and 7.06 MeV then in$ _ { 3 } \mathrm { L } ^ { 7 } + _ { 1 } \mathrm { P } ^ { 1 } \rightarrow 2 _ { 2 } \mathrm { He } ^ { 4 }$ energy released is:

  1. $29.6 \mathrm { MeV }$

  2. $2.4MeV$

  3. $8.4 \mathrm { MeV }$

  4. $17.3 \mathrm { MeV }$


Correct Option: A

Mass defect of an atom refers to 

  1. inaccurate measurement of mass of neutrons

  2. mass annihilated to produce energy to bind the nucleons

  3. packing fraction

  4. difference in the number of neutrons and protons in the nucleus


Correct Option: B
Explanation:

$ Mc^{2} + (B.E) = (N _{mN} + Z _{mP})c^{2}$
where,
         $M _{c} =  $  total mass of nucleus.
         $N _{mN} =  $  total mass of neutrons
         $N _{mP}  =  $  total mass of protons

In a fission process, nucleus A divides into two nuclei B and C, their binding energies being $\mathbf { E } _ { \mathbf { a } ^ { * } }$   $E _ { b }$  and $E _ { c }$ respectively. Ihen

  1. $\mathbf { E } _ { \mathrm { b } } + \mathrm { E } _ { \mathrm { c } } = \mathrm { E } _ { \mathrm { a } }$

  2. $\mathrm { E } _ { \mathrm { b } } + \mathrm { E } _ { \mathrm { c } } > \mathrm { E } _ { \mathrm { a } }$

  3. $\mathrm { E } _ { \mathrm { b } } + \mathrm { E } _ { \mathrm { e } } < \mathrm { E } _ { \mathrm { a } }$

  4. $\mathrm { E } _ { \mathrm { b } } \mathrm { E } _ { \mathrm { c } } = \mathrm { E } _ { \mathrm { a } }$


Correct Option: A
Explanation:

$\begin{array}{l} { E _{ b } }+{ E _{ c } }>{ E _{ a } } \ \, \because some\, \, energy\, \, is\, \, goen\, \, in\, \, breaking\, \, nuclie\, \, of\, \, A \ Hence, \ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$

For uranium nucleus. Find relation between mass and volume 

  1. $m\propto v$

  2. $m\propto \sqrt{v}$

  3. $m\propto v^2$

  4. $m\propto \dfrac{1}{v}$


Correct Option: A

The phenomenon of pair production is :

  1. The production of an electron and a positron from $\gamma$ radiation

  2. Ejection of an electron from a metal surface when exposed to ultraviolet light

  3. Ejection of an electron from a nucleus

  4. Ionization of a neutral atom


Correct Option: A
Explanation:

Pair production :
Gamma rays of sufficient energy when passing near a nucleus disappear and materialize into pair of an electron and a positron.
To have pair production, minimum energy of $\gamma$ - ray radiation is 1.02MeV.

In pair annihilation the least number of  $\gamma $- ray photons produced is :

  1. 2

  2. 3

  3. 4

  4. 1


Correct Option: A
Explanation:

In a pair annihilation, atleast two gamma rays must be produced. For example, when an electron encounters a positron, they annihilate to produce two gamma rays each having $0.511$ MeV energy.

The rest energy of electron or positron is

  1. 0.51 MeV

  2. 1 MeV

  3. 1.02 MeV

  4. 1.5 MeV


Correct Option: A
Explanation:

Mass of electron $ = 9.1 \times 10^{-31}  kg$

Rest mass energy $ = mc^{2}$


                               $ =\left (\dfrac{9.1\times 10^{-31}}{1.6\times10^{-27}} \times931.978\right )c^{2}$

                               $ = 0.52\  MeV$

Positronium is converted into

  1. 2 Photons each of energy 0.51MeV

  2. 1 Photon of energy 1.02 MeV

  3. 2 Photons each of energy 1.02MeV

  4. 1 Photon of energy 0.51MeV


Correct Option: A
Explanation:

Rest mass energy of a positron (as well as electron) is $0.51$ MeV.

Total rest mass energy of positron and electron is $2(0.51) = 1.02$ MeV
Thus a positron and an electron annihilate to produce two photons each of energy $0.51$ MeV. This phenomenon is known as positron-electron annihilation process.

In pair annihilation, two  $\gamma $ -ray photons are produced due to

  1. Law of conservation of energy

  2. Law of conservation of mass

  3. Law of conservation of momentum

  4. Law of conservation of angular momentum


Correct Option: B
Explanation:

When a particle encounters its antiparticle, they annihilate with the transformation of their mass energies into two gamma rays obeying law of conservation of mass.

To produce pair production, the minimum energy of $\gamma $-ray should be 

  1. 0.15 MeV

  2. 1 MeV

  3. 1.02 MeV

  4. 1.5 MeV


Correct Option: C
Explanation:

Minimum energy to produce pair production should be 1.02 MeV.

The energy equivalent of 1mg of mass in joule is

  1. 3 x 10$^{2}$

  2. 3 x 10$^{10}$

  3. 9 x 10$^{10}$

  4. 9 x 10$^{2}$


Correct Option: C
Explanation:

As, $m=10^{-6}\ kg$

using, $E=mc^2$
$E= 10^{-6} \ \times (3\times 10^8)^2 $
$E=9\times10^{10}J$

Which one of the following cannot be used as a moderator in a nuclear reactor?

  1. Water

  2. Heavy water

  3. Molten sodium

  4. Graphite


Correct Option: A

Which row describes the nature of $\alpha$- particles and of $\gamma$- rays

  1. $\alpha$- particles : helium nuclei ; $\gamma$ rays - electromagnetic radiation

  2. $\alpha$- particles : helium nuclei ; $\gamma$ rays - electrons

  3. $\alpha$- particles : protons; $\gamma$ rays - electromagnetic radiation

  4. $\alpha$- particles : protons; $\gamma$ rays - electrons


Correct Option: A
Explanation:
Characteristics of $\alpha$ particle
  mass= $4$ unit
  charge = $+2$ unit
The mass of Helium nuclei is $4$ unit and charge is $+2$ unit. So, it is equivalent to $\alpha $ particle .

$\gamma $- rays  is a penetrating electromagnetic radiation arising from the radioactive decay of atomic nuclei.

A scientist carries out an experiment using a sealed source which emits $\beta$ -particles. The range of the $\beta$- particles in the air is about $30cm$.
Which precaution is the most effective to protect the scientist from the radiation?

  1. handling the source with long tongs

  2. keeping the temperature of the source low

  3. opening all windows in the laboratory

  4. washing his hands before leaving the laboratory


Correct Option: A
Explanation:

Handling the source with long tongs will prevent him from the exposure of $\beta$- particles. Since the range of $\beta$-particle is only $30cm$ so, long tongs can easily be a safer and effective precaution.

One electron volt is equal to .......................

  1. $\displaystyle 1.6\times 10^{-19}$ Joule

  2. $\displaystyle 16\times 10^{-19}$ Joule

  3. $\displaystyle 1.6\times 10^{-10}$ Joule

  4. $\displaystyle 1.6\times 10^{-9}$ Joule


Correct Option: A

How much energy is released when a $ _{8}{O}^{16}$ nucleus is completely converted into energy?

The binding energy per nucleon of $ _{8}{O}^{16}$ is $7.97  MeV$ and ${m} _{p} = 1.0078  u$ and ${m} _{n} = 1.0087  u$

  1. $14899.438 MeV$

  2. $148.99 MeV$

  3. $4489.73 MeV$

  4. $448.973 MeV$


Correct Option: A
Explanation:
Binding Energy $=E=\triangle mc^2$
$\triangle m=$ mass of nucleus $-$ (mass of proton + neutron)
Multiplying both sides by $c^2$
$\triangle mc^2=$ (mass of nucleus)$c^2 -$ (mass of proton + neutron)$c^2$
$=7.97\times 16-8(1.0078+1.0087)931 \\ E=127.52-15018.892 \\ E=-14899.438MeV$

What is the energy required to increase the mass of a system by one atomic mass unit?

  1. 661.5 MeV

  2. 931.5 MeV

  3. 1336.5 MeV

  4. 785.2 MeV


Correct Option: B
Explanation:

$\Delta m = 1  u$
$E = 1 \times 931.5  MeV = 931.5  MeV$

Using $E = m{c}^{2}$, find out the energy released, when $2  u$ of mass is destroyed completely.
Take $1  u = 1.66 \times {10}^{-27}  kg$.

  1. $4.65 MeV$

  2. $3627 MeV$

  3. $91.5 MeV$

  4. $1865 MeV$


Correct Option: D
Explanation:
When $2u$ of mass is destroyed completely then
$\triangle m=2u\\ E=\triangle m{ c }^{ 2 }\\ E=2\times 931MeV\\ \therefore 1u=931Mev/{ c }^{ 2 }\\ So,E=1863MeV\approx 1865MeV$
So, (D) is correct option.
What  describe the Einsteins equation for the relativity of mass and energy ?
  1. a small amount of mass contains a lot of energy.

  2. a small amount of energy can be converted into a large amount of mass.

  3. a small amount of mass contains a small amount of energy.

  4. mass can be converted into energy, but energy cannot be converted mass.

  5. energy can be converted into mass, but mass cannot be converted into energy.


Correct Option: A
Explanation:

Sir Einstein's mass-energy equation states that mass and energy can be converted into each other by the following relation.

                 $E=mc^{2}$,    ($c=$speed of light)
This implies that a small amount of mass contains a lot of energy, which can be proved with an example.
Let we have a mass of $1g=10^{-3}kg$ , therefore energy produced by it will be:
                      $E=10^{-3}\times \left(3\times10^{8}\right)^{2}=9\times10^{13}J$ 
which is a vast amount energy produced by only one gram (small mass) of mass.
Whereas a small amount of energy doesn't give a large amount of mass because for that we have to divide the energy by $c^{2}$, which gives a small mass.

Find the accurate expressions.
  1. $E = mc^2$

  2. $E = m/c^2$

  3. $M = Ec^2$

  4. All of the expressions are accurate.

  5. $c = Em^2$


Correct Option: A
Explanation:

Before Sir Einstien, mass and energy were two completely different physical quantities, which were not related to each other anyway. Sir Einstein told that energy and mass are related to each other i.e. energy and mass can be converted into each other by the following relation:

      $E=mc^{2}$, called Sir Einstein's mass-energy equivalence

Which of the following statement is accurate for an object approaching the speed of light ?
  1. The particle's mass will increase as it approaches the speed of light.

  2. The particle's mass will increase as it approaches, and then decrease when it reaches the speed of light.

  3. The particle's mass will decrease as it approaches the speed of light.

  4. The particle's dimensions will increase but it's mass will remain constant as it approaches the speed of light.

  5. All of the statements are accurate


Correct Option: A
Explanation:

The mass of object as observed by a moving observer is given by $m=\dfrac{m _0}{\sqrt{1-(\dfrac{v}{c})^2}}$.

where $m _0$ is the rest mass of the object.

Hence, as speed approaches the speed of light, that is, as $v\rightarrow c$, $m\rightarrow \infty$

The rest energy involved in a mass of one atomic mass unit is _________ eV.

  1. $931$ MeV

  2. $1.6$ eV

  3. $9.3$ MeV

  4. $9.1$


Correct Option: A
Explanation:

931 MeV

The mass of 1 amu is equivalent to an energy of 931 MeV. 

The unit of rate constant for a zero order reaction is:

  1. $s^{-1}$

  2. $mol L^{-1}s^{-1}$

  3. $L mol ^{-1}s^{-1}$

  4. $L^{2} mol ^{-2}s^{-1}$


Correct Option: B
Explanation:
For zero order reaction
Rate $=K[A]^0$
Rate $=K=\cfrac{d[A]}{dt}$
Unit of rate constant $=Unit\; of \; \cfrac{d[A]}{dt} \\=\cfrac{mol L^{-1}}{sec} \\=mol L^{-1} sec^{-1}$

Which of the following assertions are correct?

  1. A neutron can decay to a proton only inside a nucleus

  2. A proton can change to a neutron only inside a nucleus

  3. An isolated neutron can change into proton

  4. An isolated proton can change into a neutron


Correct Option: B

Inside nucleus, protons are held together though they have the dame charge. Why?

  1. The strong attractive nuclear force far exceeds the electrostatic force between the protons

  2. Neutrons prevent them from repelling from each other

  3. The electrostatic attractive force between an electron and a proton is more than the electrostatic repulsive force between the protons

  4. Gluons are responsible for holding them together


Correct Option: A

The conversion of 1 u of mass results in ________ eV of energy.

  1. $9.315 \times 10^6$

  2. $391.5 \times 10^6$

  3. $931.5 \times 10^6$

  4. $93.15 \times 10^6$


Correct Option: C

Magnitude of mass defect is a measure of ......................... of a nucleus.

  1. Unstability

  2. Stability

  3. Charge

  4. Position


Correct Option: B

What is energy equivalent to a $10\ \mu g$ mass?

  1. $9\ \times 10^{7}\ J$

  2. $3\ \times 10^{11}\ J$

  3. $5\ \times 10^{11}\ J$

  4. $7\ \times 10^{11}\ J$


Correct Option: A

A proton and an -particle enters a uniform magnetic field moving with the same speed. If the proton Takes 25s to make 5 revolutions, what is the periodic time for the -particle?

  1. 50 s

  2. 25 s

  3. 10 s

  4. 5 s


Correct Option: D

In a laboratory experiment on emission from atomic hydrogen in a discharge tube, only a small number of lines are observed where as a lines are present in the hydrogen spectrum of a star. This is because in a laboratory  

  1. The amount of hydrogen taken is much smaller than that present in the star

  2. The temperature of hydrogen is much smaller than that of the star

  3. The pressure of hydrogen is much smaller than that of the star

  4. The gravitational pull is much larger than that in the star


Correct Option: C

Ionization energy of $Li$(Lithium) atom in ground state in $5.4 eV$. Binding energy of an electron in $Li^+$ ion in ground state is $75.6 eV$. Energy required to remove all three electrons of Lithium (Li) atom is:-

  1. $203.4 \ eV$

  2. $135.4 \ eV$

  3. $81.0 \ eV$

  4. $156.6 \ eV$


Correct Option: D

In a hypothetical star,two carbon nuclei fuse to form magnesium.The reaction is:(take :$1amu=931MeV/c^2)$
$^{12}C+^{12}C\rightarrow ^{24}Mg$
The energy released per carbon nuclei is: (Mass of $^{24}Mg=23.985amu)$

  1. $13.965 MeV$

  2. $11.12 MeV$

  3. $6.982 MeV$

  4. $10.12 MeV$


Correct Option: A
Explanation:

$^{12}C+^{12}C\rightarrow ^{24}Mg$
mass of $^{12}C=12.0642 amu$
Total input mass $m _1=2\times 12.0642$
$=24.1284$ amu
Mass of $^{24}Mg$=output mass
$=m _0=23.955$
mass loss $=m _1-m _o$
$Am=24.1284-23.985$
$=0.1434 amu$
Energy released $=0.1434\times 931 Mev$
$=13.965 Mev$

One milligram of matter converted into energy will give

  1. $9$ J

  2. $9 \times 10^{13}$ J

  3. $9 \times 10^5 $ J

  4. $9 \times ^{10}$ J


Correct Option: D

One mole of radium has an activity of 1/3.7 killo curie. Its decay constant will be 

  1. $\frac{1}{6}\times -10s^{-1}$

  2. $ 10^{-10}s^{-1}$

  3. $ 10^{-11}s^{-1}$

  4. $ 10^{-8}s^{-1}$


Correct Option: C
Explanation:

Number of mole of radium, $n=1$

The number of nuclei of radium, $N={{N} _{a}}=avagdaro\,\,number=6.023\times {{10}^{23}}$

Activity, $A=\lambda N$

$ \because 1\,\,curie=3.7\times {{10}^{10}}\,decay/\sec  $

$ \dfrac{1}{3.7}\,kilocurie=0.27\,kilocurie=9.99\times {{10}^{12}}\,dacay/\sec  $

So, $\lambda =\dfrac{A}{N}=\dfrac{9.99\times {{10}^{12}}}{6.023\times {{10}^{23}}}=1.65\times {{10}^{-11}}\,$

In each fission energy of $200\ MeV$ is released. How many acts of fission must occur per second to produce a power of $1\ kw$?

  1. $3.1\times 10^{13}$

  2. $1.3\times 10^{16}$

  3. $1.4\times 10^{16}$

  4. $2.3\times 10^{15}$


Correct Option: C

The binding energy of  $ _ { 17 } \mathrm { CI } ^ { 35 }$  nucleus is  $298\ \mathrm { MeV }.$  Find its atomic mass. The mass of hydrogen atom  $({ _{ 1 }{ H } }^{ 1 })$  is  $1.008143\  \mathrm { amu }$  and that of a neutron is  $1.008986\  \mathrm { amu }.$  Given  $1\  \mathrm { amu } = 931\  \mathrm { MeV }.$

  1. $148$

  2. $298$

  3. $340$

  4. $348$


Correct Option: D

In a working nuclear react, cadmium rods (control rods) are used to:-

  1. Speed up neutrons

  2. Slow down neutrons

  3. Absorb some neutrons

  4. Absorb all neutrons


Correct Option: C

In the nuclear reaction ; $ _{92}U^{238}\rightarrow _{z}Th^{A}+ _{2}He^{4}$ the values of A and Z are: 

  1. A=230,Z=8

  2. A=234,Z=90

  3. A=228, Z=94

  4. A=232, Z=


Correct Option: B

Out side a nucleus 

  1. Neutron is stable

  2. Proton and neuron both are stable

  3. Neutrons is unstable

  4. Neither neutrons nor proton is stable


Correct Option: C

Consider a hypothetical annihilation of a stationary electron with a stationary positron. What is the wavelength of resulting radiation?

  1. $\dfrac{h}{m _{0}c}$

  2. $\dfrac{h}{2m _{0}c}$

  3. $\dfrac{2h}{m _{0}c}$

  4. $\dfrac{h}{4\pi m _{0}c}$


Correct Option: A
Explanation:
According to De Broglie eave length
$\lambda =\dfrac { h }{ mc } \quad \quad \quad E=\dfrac { hc }{ \lambda  } $
                             $\lambda =\dfrac { hc }{ E } =hc$
When an electron and a positron collide then a hypothetical annihilation of a stationary electron with a stationary position.

The atomic mass of $7 ^ { N ^ { 15 } }$ is 15.000108 a.m.u. and that  is of $8 ^ { \bigcirc ^ { 16 } }$ 15.994915 a.m.u. If the mass of a proton is 1.007825 a.m.u. then the minimum energy provided to remove the least tightly bound proton is

  1. 0.0130181 MeV

  2. 12.13 MeV

  3. 13.018 MeV

  4. 12.13 eV


Correct Option: B
Explanation:

$\begin{array}{l} \left( { { M _{ n } }+{ M _{ H } }-{ M _{ 0 } } } \right) \times 431.5 \ =\left[ { 15.000108+1.007827-15.994915 } \right] \times 431 \ =12.13\, Mev \end{array}$

The energy of the reaction ${ Li }^{ 7 }+p\longrightarrow 2{ He }^{ 4 }$ is (the binding energy per nucleon in ${ Li }^{ 7 }$ and ${ He }^{ 4 }$ nuclei are 5.60 and 7.06 MeV respectively.)

  1. 17.3 MeV`

  2. 1.73 MeV

  3. 1.46 MeV

  4. Depends on binding energy of proton


Correct Option: D
Explanation:

The energy released will depend on the energy equivalent of proton, taking its mass to be $1amu$ or
 corresponding energy as $931.5Mev$
we get energy released as $Q=E _{reactant} -E _{product}=7\times 5.6 +931.5 -2\times4\times  7.06=914.22Mev$

The binding energy per nucleon of deuteron $(^2 _1 H)$ and helium nucleus $(^4 _2 He)$ is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

  1. 23.6 MeV

  2. 26.9 MeV

  3. 13.9 MeV

  4. 19.2 MeV


Correct Option: A

Find the  binding energy of a H atom in the state n = 2

  1. 2.1 eV

  2. 3.4 eV

  3. 4.2 eV

  4. 2.8 eV


Correct Option: B

The binding energy per nucleon of deutron $(^2 _1 H)$ and helium nucleus $(^4 _2 He)$ is 1.1 MeV and 7 MeV respectively. If two deutron nuclei react to form a single helium nucleus, then the energy released is

  1. $23.6 MeV$

  2. $26.9 MeV$

  3. $13.9 MeV$

  4. $19.2 MeV$


Correct Option: A

Binding energy per nucleon is $8.5 \text { MeV for } A = 120$ and is $7.6 \mathrm { MeV } \text { for } \mathrm { A } = 240$ Suppose a nucleus with $A = 240$ breaks into two nuclei of nearly equal mass numbers then which of the following is correct

  1. 216 MeV energy is released.

  2. 21 MeV energy is to be given from outside

  3. 220 MeV energy is released.

  4. no energy is released.


Correct Option: A

Energy released if mass of $2\ amu$ is converted into energy is :

  1. $1.5 \times 10^{-10}\ J$

  2. $3 \times 10^{-10}\ J$

  3. $1863\ J$

  4. $931.5 \Mev$


Correct Option: B
Explanation:

$ E = \Delta m c^{2}$
    $ = (2 \times 1.67 \times 10^{-27}  kg) \times (3 \times 10^{8} \frac{m}{s})^{2} $
    $ = 3 \times 10^{-10}  J$

When an electron and a positron are annihilated, then the number of photons produced is

  1. 2

  2. 1

  3. 3

  4. 4


Correct Option: A
Explanation:

Two photons are produced during the annihilation of an electron and a positron along with $1.02$ MeV released energy.

Consider the nuclear reaction: $\mathrm { X } ^ { 200 } \longrightarrow \mathrm { A } ^ { 110 } + \mathrm { B } ^ { 20 }$If the binding energy per nucleon for $\mathrm { X } , \mathrm { A }$ and $\mathrm { B }$ is $7.4 \mathrm { MeV } , 8.2 \mathrm { MeV }$ and 8.2$\mathrm { MeV }$ respectively, what is the energy relesed?

  1. $200$ $\mathrm { MeV }$

  2. $160$ $\mathrm { MeV }$

  3. $110$ $\mathrm { MeV }$

  4. $90$  $\mathrm { MeV }$


Correct Option: B

In the nucleus of helium if ${ F } _{ 1 }$ is the net force between two protons ${ F } _{ 2}$ is the net force between two neutrons and ${ F } _{ 3 }$ is the net force between a proton and a neutron. Then,

  1. ${ F } _{ 1 }={ F } _{ 2 }={ F } _{ 3 }$

  2. ${ F }> _{ 1 }{ F } _{ 2 }{ >F } _{ 3 }$

  3. ${ F }> _{ 2 }{ F } _{ 3 }{ >F } _{ 1 }$

  4. ${ F } _{3}={ F } _{ 1 }{ >F } _{ 2 }$


Correct Option: D
Explanation:

Nuclear forces are the strong forces of attraction which hold together the nucleons (neutrons and protons) in the tiny nucleus of an atom, inspite of strong electrostatic forces of repulsion between protons. Nuclear forces act between a pair of neutrons, a pair of protons and also between a neutron, proton pair with the same strength. This shows that nuclear forces are independent of charge. 
The attractive nuclear force is the same for any pair of nucleons. Thus$ F _1 = F _3$ when there are no electrostatic forces, but $F _2 =$ attractive nuclear force - repulsive electrostatic force
Hence
$F _3 = F _1 > F _2$

The binding energy of $\alpha $-particle is ( if ${ m } _{ p }=1000785$ $u,{ m } _{ n }=1.00866$ u and ${ m } _{ \alpha  }=4.00274u$)

  1. $56.42 MeV$

  2. $2.821 MeV$

  3. $28.21 MeV$

  4. $32.4 MeV$


Correct Option: C

For a pair production, the minimum frequency of the gamma ray must be:

  1. 2.5 x 10$^{14}$ Hz

  2. 2.5 x 10$^{20}$ Hz

  3. 2.5 x 10$^{28}$ Hz

  4. 2.5 x 10$^{34}$ Hz


Correct Option: B
Explanation:

For production to occur, the energy must be greater than the rest mass of proton + electron.
$E > ( 2 \times 0.511\ MeV = 1.022\ MeV)$
$h \nu > 1.022 \times  10^{6} \times 1.6 \times 10^{-19} J$
$\nu > \dfrac{1.6352 \times  10^{-13}}{h} = 2.5 \times 10^{20} Hz$

The energy released when a positron is annihilated is

  1. $0.51 MeV$

  2. $0.58 MeV$

  3. $185 MeV$

  4. $200 MeV$


Correct Option: A
Explanation:

Positron, also called positive electron,  positively charged subatomic particle having the same mass and magnitude of charge as the electron.

mass of $e^+ = 9.11\times 10^{-31}Kg$

$E =mc^2=  9.11\times 10^{-31}\times (3\times 10^8)^2  J$

                 $= {0.51  \ MeV}$

The energy of annihilation of positron is ${0.51 \ MeV}$.

If the energy of an electron in Hydrogen atom is given by expression, $-1312 /{ n }^{ 2 }kJ{ mol }^{ -1 }$, then the energy required to excite the electron from ground state to second orbit is 

  1. $328 kJ{ mol }^{ -1 }$

  2. $656 kJ{ mol }^{ -1 }$

  3. $984 kJ{ mol }^{ -1 }$

  4. $1312 kJ{ mol }^{ -1 }$


Correct Option: C
Explanation:

$\begin{array}{l} E=-1312\left( { \frac { 1 }{ 4 } -1 } \right)  \ =\frac { { -1312\times 3 } }{ 4 }  \ =328\times 3 \ =984\, kJmo{ l^{ -1 } } \ Hence, \ option\, \, C\, \, is\, \, correct\, answer. \end{array}$

$\gamma $ -ray photon of following energy undergoes pair production : 

a) $0.85Mev $      b) $1.00Mev $
c)  $1.02Mev $     d) $1.82Mev$

  1. a,b

  2. c,d

  3. b,c,d

  4. only d


Correct Option: B
Explanation:

The pair production becomes possible with gamma energies exceeding $1.02\ MeV$, and becomes important as an absorption mechanism at energies over $5\ MeV$.

Assertion (A) : Due to annihilation of electron positron pair, at least 2 $\gamma $-ray photons are produced.
Reason (R) : This is in accordance with conservation of linear momentum.

  1. Both A & R are true and R is the correct explanation of A

  2. Both A & R are true and R is not correct explanation of A

  3. A is true but R is false

  4. A is false but R is true


Correct Option: A
Explanation:

$ e^{-} + e^{+} \Rightarrow  \gamma +\gamma $
This annihilation is in accordance to three laws:-
1] conservation of linear momentum
2] conservation of charge
3] conservation of angular momentum
Two $\gamma $ ray photons are released on annihilation of electron and positron.

Choose the correct statement :

  1. A nucleus is relatively more stable for which total binding energy is more.

  2. A nucleus is relatively more stable for which binding energy per nucleon is more.

  3. A nucleus is relatively more stable for which total binding energy is low.

  4. None of these


Correct Option: A
Explanation:

Binding energy is the amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals. Binding energy is more if the nucleus is relatively more stable.

In the nuclear reaction : $X(n, \alpha) _3 Li ^7$ the term X will be 3

  1. $ _5{B}^{10}$

  2. $ _5{B}^{9}$

  3. $ _5{B}^{11}$

  4. $ _2{He}^{4}$


Correct Option: B

The energy equivalent to $1kg$ of matter in (in Joule)

  1. $10^{17}$

  2. $19^{20}$

  3. $10^{11}$

  4. $10^{14}$


Correct Option: A
Explanation:

Einstein mass -energy relation : $E=mc^2=1\times (3\times 10^8)^2=9\times 10^{16} \sim 10^{17}$ J

$1$ $a.m.u$ is equivalent to 

  1. $931$ $MeV$

  2. $139$ $MeV$

  3. $93$ $MeV$

  4. $39$ $MeV$


Correct Option: A
Explanation:

$m=1\ a.m.u.=1.66\times10^{-27}kg$

Its energy equivalent, using $E=mc^2$ is $E=1.49\times10^{-10}\ J$.
or $E=1.49\times10^{-10}/1.6\times10^{-13}\ MeV=931\ MeV$

The binding energy per nucleon of deuteron $ \left( \frac { 2 }{ 1 }  H \right)  $ and helium nucleus $ \left( \frac { 4 }{ 2 } He \right)  $ is.1.1 meV and 7 meV respectively. If two deuteron nuclei react to from s single helium nucleus, then the energy released is:

  1. 13.9 MeV

  2. 26.9 MeV

  3. 23.6 MeV

  4. 19.2 MeV


Correct Option: C

The energy equivalent to a substance of mass $1$g is?

  1. $18\times 10^{13}$J

  2. $9\times 10^{13}$J

  3. $18\times 10^6$J

  4. $9\times 10^6$J


Correct Option: B
Explanation:

$E=mC^2$

$=1\times 10^{-3}\times 9\times 10^{16}$
$=9\times 10^{13}$J

The binding energy expressed in $MeV$ is given for the following nuclear reactions :
$ _2He^3+\ _0n^1\rightarrow\ _2He^4+20\ MeV$
$ _2He^4+\ _0n^1 \rightarrow\ _2He^5 -0.9\ MeV$
Which of the following conclusions are correct ?

  1. $ _2He^4$ is less stable than both $ _2He^3$ and $ _2He^5$

  2. $ _2He^4$ is less stable than $ _2He^3$ but more stable than $$ _2He^5$

  3. $ _2He^4$ is less stable than $ _2He^5$ but more stable than $$ _2He^3$

  4. $ _2 He^4$ is more stable than both $ _2He^3$ and $ _2He^5$


Correct Option: D

An electron and a positron are moving side by side in the positive $x-$direction at $1.5\times 10^8\ m/s$. When they annihilate each other, two photons are produced that move along the $x-$axis, then : 

  1. both move in positive $x-$direction

  2. both move in positive direction along $x-$axis

  3. both may move in same direction

  4. both $(a)$ and $(c)$ are correct


Correct Option: B

In nuclear reaction
$ _{2}He^{4}+\ _{Z}X^{A}\rightarrow Z+\  _{2}\gamma^{A+3}+\ _{Z}M^{A}$
where $M$ denotes

  1. electron

  2. positron

  3. proton

  4. neutron


Correct Option: D

One milligram of matter converted into energy will give:

  1. $90\ J$

  2. $9\times 10^{3}\ J$

  3. $9\times 10^{10}\ J$

  4. $9\times 10^{3}\ J$


Correct Option: C

A parent nucleus $^{m} _{1}p$ decays into a daughter nucleus $D$ through $\alpha$ emission in the following way $^{m} _{1}p\rightarrow D+\alpha$ The subscript and superscript on the daughter nucleus $D$ will be written as

  1. $^{m} _{n}D$

  2. $^{m+4} _{n}D$

  3. $^{m-4} _{n}D$

  4. $^{m-4} _{n-2}D$


Correct Option: D

Name the following nuclear reaction :
$ _{92}U^{238}(\alpha, 6p, 13n) _{88}Ra^{228}$

  1. particle-particle reaction

  2. capture reaction

  3. fission reaction

  4. separation


Correct Option: D

Consider a hypothetical annihilation of a stationary electron with a stationary positron. What is the wavelength of the resulting radiation?

  1. $\lambda = \displaystyle\frac{h}{2m _ec}$

  2. $\lambda = \displaystyle\frac{2h}{m _ec^2}$

  3. $\lambda = \displaystyle\frac{h}{2m _ec^2}$

  4. None of these


Correct Option: A
Explanation:

Annihilation is the conversion of matter into energy.
Using equation $E={ m } _{ e }{ c }^{ 2 }$
$\dfrac { hc }{ \lambda  } =2{ m } _{ e }{ c }^{ 2 }$
$\Rightarrow \lambda =\dfrac { h }{ { 2m } _{ e }{ c } } $

In which of the following nuclear reactions, the product is incorrectly matched ?

  1. $ _{96}Cm^{242}(\alpha, 2n) _{97}Bk^{243}$

  2. $ _5B^{10}(\alpha, n) _7N^{13}$

  3. $ _7N^{14}(n, p) _6C^{14}$

  4. $ _{14}Si^{28}(d, n) _{15}P^{29}$


Correct Option: A

A proton and an alpha particle having same momentum enter a magnetic field at right angles to it. If $r _1$ and $r _2$ be their radii respectively then value of $r _1 /r _2$ is : 

  1. $1$

  2. $2$

  3. $1/2$

  4. $1/4$


Correct Option: B

$A5\times 10^{-4}\overset {o}{A}$ photon produces an electron-positron pair in the vicinity of a heavy nucleus. Rest energy of electron is 0.5 11 MeV. If they have the same kinetic energies, the energy of each particle is nearly

  1. 1.2 MeV

  2. 12 MeV

  3. 120 MeV

  4. 1200 MeV


Correct Option: B

If mass-energy equivalence is taken into account, when water is cooled to form ice, the mass of water should

  1. increase

  2. remain unchanged

  3. decrease

  4. first increase then decrease


Correct Option: A
Explanation:

Because thermal energy decreases, therefore mass should increase.

If 1mg of ${ U }^{ 235 }$ is completely annihilated, the energy liberated is

  1. $\quad 9\times { 10 }^{ 10 } J$

  2. $\quad 9\times { 10 }^{ 19} J$

  3. $\quad 9\times { 10 }^{ 18} J$

  4. $\quad 9\times { 10 }^{ 17} J$


Correct Option: A
Explanation:

$E={ mc }^{ 2 }={ 10 }^{ -6 }\times \left( 3\times { 10 }^{ 8 } \right) ^{ 2 }={ 10 }^{ -6 }\times 9\times { 10 }^{ 16 }=9\times { 10 }^{ 10 }J$

One milligram of matter convert into energy will give

  1. $90 joule$

  2. $9\times { 10 }^{ 3} joule$

  3. $9\times { 10 }^{ 5} joule$

  4. $9\times { 10 }^{ 10} joule$


Correct Option: D
Explanation:

$E={ mc }^{ 2 }={ 10 }^{ -3 }\times{ 10 }^{ -3 } \left( 3\times { 10 }^{ 8 } \right) ^{ 2 }=9\times { 10 }^{ 10 }J$

The mass and energy equivalent to $1 amu$ are respectively

  1. $1.67\times { 10 }^{ -27 }gm$, $9.30 MeV$

  2. $1.67\times { 10 }^{ -27 }kg$, $930 MeV$

  3. $1.67\times { 10 }^{ -27 }kg$,$ 1 MeV$

  4. $1.67\times { 10 }^{ -34 }kg$, $1 MeV$


Correct Option: B
Explanation:

1 amu
It is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.


Mass of 1 mole C atoms $= 12 gms$
$1\ amu = $$ \cfrac{12}{12 \times 6.023 \times 10^{23}} = 1.66 \times 10^{-27} kg$

Energy equivalent, $E = mc^2 =1.66 \times 10^{-27} kg \times (3 \times 10^8)^2$$ = 930\ MeV$

If an electron and positron annihilate, then the energy released is

  1. $3.2\times { 10 }^{ -13 } J$

  2. $1.6\times { 10 }^{ -13 } J$

  3. $4.8\times { 10 }^{ -13 } J$

  4. $6.4\times { 10 }^{ -13 } J$


Correct Option: B
Explanation:

Energy of electron $={ m } _{ e }{ c }^{ 2 }$
Energy of positron $={ m } _{ p }{ c }^{ 2 }$
${ m } _{ e }$ = ${ m } _{ p }$, $ c=$ speed of light.

Thus according to conservation of energy released $=2{ m } _{ e }{ c }^{ 2 }$$=2\times 9.1\times { 10 }^{ -31 }\left( 3\times { 10 }^{ 8 } \right) ^{ 2 }=1.6\times { 10 }^{ -13 } Joules.$

The rest energy of an electron is

  1. $510 KeV$

  2. $931 KeV$

  3. $510 MeV$

  4. $931 MeV$


Correct Option: A
Explanation:

Rest energy of an electron $=m _{ e }{ c }^{ 2 }$
Here $m _{ e }=9.1\times { 10 }^{ -31 }kg$ and $C$ = velocity of light

$\therefore$ Rest energy $=9.1\times { 10 }^{ -31 }\times\left( 3\times { 10 }^{ 8 } \right) ^{ 2 }joule$

                         $=\displaystyle \frac { 9.1\times { 10 }^{ -31 }\times \left( 3\times { 10 }^{ 8 } \right) ^{ 2 } }{ 1.6\times { 10 }^{ -19 } } eV\simeq 510KeV$

1mg of matter convert into energy will give

  1. $90$ joule

  2. $9\times { 10 }^{ 3 }$ joule

  3. $9\times { 10 }^{ 5 }$ joule

  4. $9\times { 10 }^{ 10 }$ joule


Correct Option: D
Explanation:

$E=m{ c }^{ 2 }=10^{ -3 }\times { 10 }^{ -3 }\times { \left( 3\times { 10 }^{ 8 } \right)  }^{ 2 }= 9\times { 10 }^{ 10 }J$

The mass defect in a particular nuclear reaction in 0.3 grams.The amount of energy liberated in kilowatt hour is $\left( Velocity\ of \  light=3\times { 10 }^{ 8 }m/s \right) $

  1. $1.5\times { 10 }^{ 6 }$

  2. $2.5\times { 10 }^{ 6 }$

  3. $3\times { 10 }^{ 6 }$

  4. $7.5\times { 10 }^{ 6 }$


Correct Option: D
Explanation:

Mass defect in a nuclear reaction          $\Delta M = 0.3    g  =  3 \times 10^{-4}    kg$

Thus amount of energy released        $E = \Delta M    c^2  =  (3 \times 10^{-4}) \times (3 \times 10^8)^2           J$
$\implies         E =  27  \times 10^{12}      J                                  (1   kWh = 3.6  \times 10^6    J)$
$\therefore         E  = 7.5   \times 10^{6}     kWh $

The binding energy per nucleon for $\displaystyle { C }^{ 12 }$ is $7.68 MeV$ and that for $\displaystyle { C }^{ 13 }$ is $7.5 MeV$. How much energy is  required to remove a neutron from $\displaystyle { C }^{ 13 }$ ?

  1. $5.34MeV$

  2. $5.5MeV$

  3. $9.5 MeV$

  4. $9.34MeV$


Correct Option: A
Explanation:

Total B.E. for $C^{13}$ is $13\times7.5=97.5\ MeV$

Total B.E. for $C^{12}$ is $12\times7.68=92.16\ MeV$
The energy required to remove a neutron from $C^{13}$ is $97.5-92.16=5.34\ MeV$

When a neutron collides with a quasi free proton, it loses half of its energy on the average in the every collission. How many collisions, on the average, are required to reduce a 2 MeV neutron to a thermal energy df 0.04 eV.

  1. 30

  2. 22

  3. 35

  4. 26


Correct Option: D
Explanation:

Let $E _{0}$ be the initial energy of neutron, the energy of neutron after 1 collision reduces to $E _{0}/2=E _{1}(let)$ i.e. $E _{1}/E _{0}=1/2$. 


After second collision, $E _{2}/E _{0}=(1/2)^{2}$, therefore after $n$ collision.
           $\dfrac{E _{n}}{E _{0}}=(\dfrac{1}{2})^{n}$ 


Here, given $E _{0}=2MeV , E _{n}=0.04eV=0.04\times10^{-6}MeV$ 

Hence, $\dfrac{0.04\times10^{-6}}{2}=(\dfrac{1}{2})^{n}$ 

             $2\times10^{-8}=(\dfrac{1}{2})^{n}$ 

             $log 2-8log10=-nlog2$ 

             $0.3010-8=-0.3010n$ 

             $n=0.7699/0.3010=25.58$

Find the energy released during the following nuclear reaction.


$ _{1}{H}^{1}  +   _{3}{Li}^{7}  \longrightarrow   _{2}{He}^{4}  +   _{2}{He}^{4}$

The mass of $ _{3}{Li}^{7}$ is $7.0160  u$,  $ _{2}{He}^{4}$ is $4.0026  u$ and proton is $1.0078  u$.

  1. 19.285  MeV

  2. 14.232 MeV

  3. 17.326 MeV

  4. 23.564 MeV


Correct Option: C
Explanation:

The mass of the reactant nuclei $= 7.0160 + 1.0078 = 8.0238  u$
The mass of the product nuclei $= 4.0026 + 4.0026 = 8.0052  u$
Mass defect $= \Delta m = 8.0238 - 8.0052 = 0.0186  u$
Energy released $= 0.0186  u \times 931.5  MeV = 17.326  MeV$

The binding energy of $ _{3}{Li}^{7}$ and $ _{2}{He}^{4}$ are $39.2  MeV$ and $28.24  MeV$ respectively. Which of the following statements is correct?

  1. Helium is more stable than lithium.

  2. Lithium is more stable than helium.

  3. Both are equally stable

  4. None of the above


Correct Option: A
Explanation:

The nucleons present in $ _{3}{Li}^{7}$ is $7$.
The binding energy per nucleon for lithium is ${39.2}/{7} = 5.6  MeV$
The binding per nucleon for helium is ${28.24}/{4} = 7.06  MeV$
The binding energy per nucleon is the measure of stability of the nuclei. Therefore, helium is more stable than lithium.

Katen was studying nuclear physics. There, he collected values of binding energies of $ _{1}{H}^{2},   _{2}{He}^{4},   _{26}{Fe}^{56}$ and $ _{92}{U}^{235}$ and they are $2.22  MeV,  28.3  MeV,  492  MeV$ and $1786  MeV$ respectively. Then, he got a doubt that stability of the nucleus depends on its binding energy, which among the above four is the most stable nucleus?

  1. ${He} _{2}^{4}$

  2. ${U} _{92}^{235}$

  3. $ _{1}{H}^{2}$

  4. $ _{26}{Fe}^{56}$


Correct Option: D
Explanation:
Stability of nucleus $\alpha$ $\cfrac{Binding\;Energy}{Atomic\;mass}$
So, ${ _{ 1 }{ H }^{ 2 } }\rightarrow \cfrac { 2.22 }{ 2 } =1.11,\quad { _{ 2 }{ He }^{ 4 } }\rightarrow \cfrac { 28.3 }{ 4 } =7.075\\ { _{ 26 }{ Fe }^{ 56 } }\rightarrow \cfrac { 492 }{ 56 } =8.7,\quad { _{ 92 }{ U }^{ 235 } }\rightarrow \cfrac { 1786 }{ 235 } =7.6$
So, ${ _{ 26 }{ Fe }^{ 56 } }$ is stable among all four.

In the nuclear reaction, there is a conservation of ______.

  1. momentum

  2. mass

  3. energy

  4. all of these


Correct Option: A
Explanation:

In a nuclear reaction, there may be conversion of some mass into energy. So,both mass and energy are not conserved. It is the momentum which is conserved.a

The difference between a nuclear reactor and an atomic bomb is that

  1. no chain reaction takes place in nuclear reactor while in the atomic bomb there is a chain reaction

  2. the chain reaction in nuclear reactor is controlled

  3. the chain reaction in nuclear reactor is not controlled

  4. no-chain reaction takes place in atomic bomb while it takes place in nuclear reactor


Correct Option: B
Explanation:

The chain reaction in nuclear reactor is controlled 

Both in nuclear reactor and atomic bomb nuclear fission takes place. But in nuclear reactor controlled fission chain reaction takes place while in atomic bomb chain reaction is uncontrolled. 

The energy equivalent of $1\ amu$ is

  1. $931\ eV$

  2. $93.1\ V$

  3. $931\ MeV$

  4. $9.31\ MeV$


Correct Option: C
Explanation:

$1\ amu =1.66\times 10^{-27} kg$


According to Einstein's mass energy equivalence, $E=mc^2$ where $c=$ velocity of light. 

So, $E=1.66\times 10^{-27}\times (3\times 10^8)^2=14.94\times 10^{-11} J$

$E=\dfrac{14.94\times 10^{-11}}{1.6\times 10^{-19}} eV$       where $1eV=1.6\times 10^{-19} J$

$E=931\times 10^{6} eV=931\ MeV$

The binding energy per nucleon of $^{16}O$ is $7.97MeV$ and that of $^{17}O$ is $7.75MeV$. The energy in MeV required to remove a neutron from $^{17}O$ is:

  1. $3.52$

  2. $3.64$

  3. $4.23$

  4. $7.86$

  5. $1.68$


Correct Option: C
Explanation:

BE per nucleon $^{16}O=7.97MeV$
BE per nucleon $^{17}O=7.75MeV$
$^{17}O\rightarrow { _0n^1}+{^{16}O}$
Energy required to remove neutron
$=17\times 7.75-16\times 7.97$
$=4.23MeV$.

The mass defect of a certain nucleus is found to be $0.03$ amu. Its binding energy is:

  1. $27.93$ eV

  2. $27.93$ keV

  3. $27.93$ MeV

  4. $27.93$ GeV


Correct Option: A
Explanation:

Mass defect  $\Delta M = 0.03$ amu
Binding energy  $E _{B} = \Delta Mc^2 = \Delta M\times 931.5 $  MeV
$\therefore \ E _{B} = 0.03\times 931.5$ MeV $ =27.93 $ MeV

Consider the following statements
(i)All isotopes of an element have the same number of neutrons
(ii)Only one isotope of an element can be stable and non -radioactive
(iii)All elements have isotopes  
(iv)All isotopes of Carbon can form chemical compounds with Oxygen -16
The correct option regarding an isotope is 

  1. (iii) and (iv) only

  2. (ii),(iii) and (iii) only

  3. (i),(ii) and (iii) only

  4. (i),(iii) and (iv) only


Correct Option: A
Explanation:

A

Isotopes are atoms of the same element having the same numbers of protons and different numbers of neutrons. All elements have isotope. Also all isotope of carbon can form chemical compounds with oxygen - 16.

Higher the mass defect, higher will be the stability of the nucleus.

  1. True

  2. False


Correct Option: B
Explanation:
Amount of energy required to break the nucleus is known as binding energy of the nucleus.
It depends upon mass defect.
Greater  the mass defect, greater will be the binding energy
Stable nuclei have high binding energies

1 u is equivalent to an energy of

  1. 9.315 MeV

  2. 931.5 KeV

  3. 93.15 MeV

  4. 931.5 MeV


Correct Option: D
Explanation:
According to Einstein mass energy equivalence is represented by
$E=m{ c }^{ 2 }$
Taking $m=1a.m.u$
$=1.66\times { 10 }^{ -27 }㎏$
and $c=3\times { 10 }^{ 8 }㎧$
We get, $E=1.66\times { 10 }^{ -27 }\times { \left( 3\times { 10 }^{ 8 } \right)  }^{ 2 }J$
$=1.49\times { 10 }^{ -10 }J$
As $1MeV=1.6\times { 10 }^{ -13 }J$
$\therefore E=\cfrac { 1.49\times { 10 }^{ -10 } }{ 1.6\times { 10 }^{ -13 } } $
$E=931.25MeV$
Hence,$1a.m.u.=931.25MeV$

The mass equivalent of 931.5 MeV energy is

  1. $1.66 \times 10^{-27} kg $

  2. $6.02 \times 10^{-24}kg$

  3. $1.66 \times 10^{-20} kg$

  4. $6.02 \times 10^{-27} kg$


Correct Option: A
Explanation:
According to Einstein, mass energy equivalence is represented by
$E=m{ C }^{ 2 }$
Taking $m=1a.m.u$
$=1.66\times { 10 }^{ -27 }㎏$
and $C=3\times { 10 }^{ 8 }㎧$
We get $E=1.66\times { 10 }^{ -27 }\times { (3\times { 10 }^{ 8 }) }^{ 2 }J$
$=1.49\times { 10 }^{ -10 }J$
As $1MeV=1.6\times { 10 }^{ -13 }J$
$\therefore E=\cfrac { 1.49\times { 10 }^{ -10 } }{ 1.6\times { 10 }^{ -13 } } $
$E=931.25MeV$
Hence the mass equivalent of $931.25MeV$ energy is $1.66\times { 10 }^{ -27 }㎏$

If mass-energy equvalence is taken into account, when water is cooled to form ice, the mass of ater should

  1. Increase

  2. Remain unchanged

  3. decrease

  4. First increase and then decrease


Correct Option: C
Explanation:

As we know that when water is converted into ice then it releases 

energy to the atmosphere, in other words, it can be said that
 it releases some mass to the atmosphere because mass and 
energy are equivalent. 
So option C is correct that mass will decrease.

Two light nuclei of masses $m _1$ and $m _2 $ are fused to form a more stable nucleus of mass $m _3$ then :-

  1. $m _3 = | m _1 - m _2 | $

  2. $m _3 < ( m _1 + m _2 ) $

  3. $m _3 > ( m _1 - m _2 ) $

  4. $m _3 = | m _1 + m _2 | $


Correct Option: B
Explanation:

When two nuclei of masses ${m _1}$ and ${m _2}$ are fused to form a stable nucleus of mass ${m _3}$ and some of the mass is converted in energy.

Therefore,

${m _3} < {m _1} + {m _2}$

A photon of $1.7 \times 10 ^{-13}$ joule is absorbed by a material under special circumstances. The correct statement is :

  1. Electron of the atoms of absorbed material will go the higher energy states.

  2. Electron and positron pair will be created

  3. Only positron pair will be produced

  4. Photoelectric effect will occur and electron will be produced


Correct Option: B
Explanation:

For electron and positron pair production, minimum energy is $1.02\ MeV$.
Energy of photon is given: $ 1.7 \times 10^{-3} J=\dfrac{1.7 \times 10^{-13}}{1.6 \times 10^{-19}}$$=1.06 \, MeV$.
Since energy of photon is greater than 1.02 MeV, electron positron pair will be created.

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