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Beta decay - class-XI

Description: beta decay
Number of Questions: 66
Created by:
Tags: nuclei atomic, nuclear and particle physics radioactivity atomic nuclei nuclear physics physics
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A radioactive element ${X} _{90}^{238}$ decays into ${Y} _{83}^{222}$. The number of $\beta$-particles emitted are

  1. 1

  2. 2

  3. 4

  4. 6


Correct Option: A
Explanation:

The radioactive element undergoes four alpha decays and one beta decay as follows:


$^{238} _{90}X\rightarrow ^{222} _{83}Y+4^4 _2He+ ^{0} _{-1}e$

Clearly only one beta particle (electron) is emitted.

The mass number of an element in a radioactive series is 223. Then the radioactive series is ................

  1. 4n

  2. 4n+3

  3. 4n+2

  4. 4n+1


Correct Option: B

A radio isotope X has a half life of $10s$. Find the number of active nuclei in the sample (if initally there are $1000$ isotopes which are falling from rest from a height of $3000m$) when it is at a height of $1000m$ from the reference plane: 

  1. $50$

  2. $250$

  3. $29$

  4. $100$


Correct Option: B
Explanation:

Time taken in falling a height $h=3000-1000=2000m$ 

is given as $t=\sqrt[2]{\dfrac{2h}{g}}$
putting $g=10,h=2000$ we get $t=20second$
number of half life in this time period is $n=20/10=2$
So number of active nuclei$ = initial/2^n=initial/2^2=inital/4=1000/4=250$
Option B is correct.

When a $\beta^-$ particle is emitted from a nucleus, the neutron-proton ratio:

  1. is decreased

  2. is increased

  3. remains the same

  4. first (A) then (B)


Correct Option: B
Explanation:

$ _{A}^{Z}\textrm{X}$ $\rightarrow  _{A-1}^{Z}\textrm{Y} $  $+  \beta^{-1}$


So,  the neutron-proton ratio before emission $ = \dfrac{Z-A}{A}$

And, the neutron-proton ratio after emission $ = \dfrac{Z-A+1}{A-1}$
Since, $ \dfrac{Z-A+1}{A-1}$  $ >\dfrac{Z-A}{A}$
Therefore, B is correct option.

A certain mass of an ideal diatomic gas contained in a closed vessel is heated. It is observed that half the amount of gets dissociated, but the temperature remains constant. The ratio of the heat supplied to the gas to the initial internal energy of the gas will be

  1. $1:2$

  2. $1:4$

  3. $1:5$

  4. $1:10$


Correct Option: A

A positron is emitted by radioactive nucleus of proton number $90$. The product nucleus will have proton number :

  1. $91$

  2. $90$

  3. $89$

  4. $88$


Correct Option: C
Explanation:

The given nuclear reaction is-

$ _{90}X \rightarrow   _{+1}e $  $+ $ $ _zY$ 
Using law of conservation of atomic (or proton) number:
$90 = 1 + Z$
$\implies$ $Z = 89$
Thus the product nucleus will have proton number $89$.

When $ _{15}P^{30}$ decays to become  $ _{14}Si^{30}$, which particle is released ?

  1. electron

  2. $\alpha$-particle

  3. neutron

  4. positron


Correct Option: D
Explanation:

The nuclear reaction :   $ _{15}P^{30}\rightarrow$   ${14}Si^{30} + $  $ _{+1}e^0$

Thus a positron is emitted during the decay of  $ _{15}P^{30}$ into   $ _{14}Si^{30}$.

A nucleus $ _{  }^{ 220 }{ X }$ at rest decays emitting an $\alpha$- particle. If energy of daughter nucleus is $0.2MeV$, $Q$ value of the reaction is

  1. $10.8MeV$

  2. $10.9MeV$

  3. $11MeV$

  4. $11.1MeV$


Correct Option: A

The antiparticle of electron is

  1. positron

  2. $\alpha $-particle

  3. proton

  4. $\beta $-particle


Correct Option: A
Explanation:

Antiparticle of electron $ _{ -1 }{ e }^{ 0 }$ is positron $ _{ +1 }{ e }^{ 0 }$

Which decay increases the atomic number?

  1. Alpha decay

  2. ${\beta}^{-}$ decay

  3. ${\beta}^{+}$ decay

  4. Electron capture

  5. Gamma decay


Correct Option: B
Explanation:

$\beta^-$ decay : $^A _ZX \rightarrow ^A _{Z+1}Y $ $+ $  $ _{-1}^0e $

Thus, the atomic number is increased in $\beta^-$ decay.

What would be an atom that has lost an electron?

  1. Positron

  2. Negatron

  3. Baryon

  4. Hadron

  5. Ion


Correct Option: E
Explanation:

We know that an atom is electrically neutral because it has equal number of electrons(negative charge) and protons(positive charge) , now when it has lost an electron , negative charge decreases in the atom hence the net charge is now positive, the atom would be an ion.

For which of the following events will the resulting products have more mass than the mass of the stuff from which the products came?

  1. Alpha decay

  2. Beta decay

  3. An exothermic nuclear reaction

  4. An endothermic nuclear reaction

  5. Nuclear fission of uranium $-235$


Correct Option: D
Explanation:

In an endothermic nuclear reaction, heat energy is given to the reaction to proceed and thus according to Einstein's mass-energy equivalence law, heat energy given gets converted into additional mass which results in the formation of products with more mass than the reactants.

The equation $ _{88}Ra^{226}\rightarrow _{86}Rn^{222}+ _{2}He^{4}$ emits which particle?

  1. $\beta$-decay

  2. $\alpha$-decay

  3. $\gamma$-decay

  4. None of the above


Correct Option: B
Explanation:

$\alpha $ - decay 

When an unstable atomic nucleus emits two protons and two neutrons the radioactive process is known as alpha decay. 

During $\beta^-$ emission:

  1. a neutron in the nucleus decays emitting an electron

  2. an atomic electron is ejected

  3. an electron already present within the nucleus is ejected

  4. a part of the binding energy of the nucleus is converted into an electron

  5. a proton in the nucleus decays emitting an electron


Correct Option: A
Explanation:

$\beta^-$ emission is due to decay of neutron in the nucleus $n\rightarrow p+e^-$.

Nuclei of a radioactive element $A$ are being produced at a constant rate $\alpha$. The element has a decay constant $\lambda$. At $t =0$, there are $N _{0}$ nuclei of the element.
If $\alpha = 2N _{0}\lambda$, calculate the number of nuclei of $A$ after one half life of $A$, and also the limiting value of $N$ as $t\rightarrow \infty$.

  1. $\dfrac {4N _{0}}{2}, 2N _{0}$.

  2. $\dfrac {3N _{0}}{2}, 2N _{0}$.

  3. $\dfrac {5N _{0}}{2}, 2N _{0}$.

  4. $\dfrac {6N _{0}}{2}, 2N _{0}$.


Correct Option: B

$90$% of a radioactive sample is left undecayed after time $t$ has elapsed. What percentage of the intial sample will decay in a total time $2t$:

  1. $20$%

  2. $19$%

  3. $40$%

  4. $38$%


Correct Option: B

$ _{84}P _{0}^{210}$ originally at rest emits $\alpha $- particles of KE 'K' Find  the KE of recoiling nucleus:

  1. $\dfrac{4}{214}K$

  2. $\dfrac{4}{206}K$

  3. $\dfrac{K}{206}$

  4. $\dfrac{K}{214}K$


Correct Option: B
Explanation:

$84P _o^{210}\xrightarrow 82Pb^{206}+\alpha \Rightarrow $ mass of remaining nucleus $(Pb)=206$

Initial momentum =$0$ (initial Po was at rest)
Momentum of $\alpha$- particle carrying $KE=K$
$P _1\sqrt{2m _{\alpha}k}$     $m _{alpha}=4n$,mass of $\alpha$
$P _1=\sqrt{8k}$
from conservation of momentum
$P _1+P _2=0$
$\Rightarrow P _1^2=P _2^2$
$2\ mrk'=2k$
$K'=\dfrac{4k}{mr}=\dfrac{4k}{206}$

When a radioactive nucleus emits a $\beta $- particular, the proton- neutron ratio

  1. decreases

  2. increases

  3. remains same

  4. first decreases and increases


Correct Option: B
Explanation:

When the neutron to proton ratio in the nucleus is too great a beta particle is emitted. In basic beta decay a neutron is transformed into a proton and an electron. The electron is then emitted as a beta particle which increases the atomic number by one and the molar mass is unchanged.

Hence, the proton- neutron ratio increase.

A free neutron is unstable against $\beta$ decay with a half life of about $600$ seconds:

  1. The expression of this decay process in $n\rightarrow p+e^{-}+\vec{v}$

  2. If three are $600$ free neutrons initially, the time by which $450$ of them have decayed is $2400$ sec.

  3. The dacay rate of the sample is $0.593$ Bq.

  4. The dacay rate of the sample is $593$ Bq.


Correct Option: C

Initial number of nuclei of a radioactive substance is $5 \times 10 ^ { 16 }$ and half-life is $10$ yrs. Find the number of nuclei decayed in $5$ yrs.

  1. $2 \times 10 ^ { 16 }$

  2. $1.5 \times 10 ^ { 16 }$

  3. $3.5 \times 10 ^ { 16 }$

  4. $2.5 \times 10 ^ { 16 }$


Correct Option: B
Explanation:

$Remaining$ nuclei after $5$ years will be $N=5\times 10^{16} \times (\dfrac{1}{2})^{5/10}=\dfrac{5\times 10^{16}}{\sqrt[2]{2}}=\dfrac{5\times 10^{16}}{1.414}=3.54\times 10^{16}$


So the decayed nuclei will be $(5-3.54)\times 10^{16}=1.46\times 10^{16}$
nearly $1.5\times 10^{16}$

A mixture consists of two radioactive materials ${ A } _{ 1 }$ and ${ A } _{ 2 }$ with half lives of 20 s and 10 s respectively. Initially the mixture has $40 g$ of ${ A } _{ 1 }$ and $160 g$ of ${ A } _{ 2 }$. The active amount of the two in the mixture will become equal after :

  1. $20s$

  2. $40s$

  3. $60s$

  4. $80s$


Correct Option: B

Samples of two radioactive nuclides $A$ and $B$ are taken. $\lambda _ { A }$ and $\lambda _ { B }$ are the disintegration constants of $A$ and $B$ respectively. In which of the following cases, the two samples can simultaneously have the same decay rate at any time ? 

  1. Initial rate of decay of $A$ is twice the initial rate of decay of $B$ and $\lambda _ { A } = \lambda _ { B }$

  2. Initial rate of decay of $A$ is twice the initial rate of decay of $B$ and $\lambda _ { A } > \lambda _ { B }$

  3. Initial rate of decay of $B$ is twice the initial rate of decay of $A$ and $\lambda _ { A } > \lambda _ { B }$

  4. Initial rate of decay of $B$ is same as the rate of decay of $A$ at t = 2h and $\lambda _ { B } < \lambda _ { A }$


Correct Option: A
Explanation:

$\begin{array}{l} N={ N _{ 0 } }{ e^{ -\lambda t } } \ \therefore if\, initial\, rate\, is\, same\, \, and\, { \lambda _{ A } }={ \lambda _{ B } } \end{array}$

Hence,
option $(A)$ is correct answer.

The radius of spherical nucleus as measured by electron scattering is 36. fm. what is the likely mass number of the nucleus?

  1. 27

  2. 40

  3. 56

  4. 120


Correct Option: A

A radioactive element $ _ { 90 } \mathrm { X } ^ { 238 }$ decays into $\mathrm { 83 } \mathrm { Y } ^ { 222 }$,then the number of $\beta$ -particles emitted are

  1. $1$

  2. $2$

  3. $4$

  4. $6$


Correct Option: C

A bone containing 200 g carbon-14 has a $\beta $ decay rate of 375 deacy/min. Calculate the time that has elapsed since the death of the living one. Given the rate of decay for the living organism is equal to 15 decay per min per gram of carbon and half - life of carbon -14 is 5730 years,

  1. 27190 years

  2. 1190 years

  3. 17190 years

  4. None of these


Correct Option: A

A nucleus X undergoes following transformation
$x \stackrel { a } { \longrightarrow } Y$
$Y \longrightarrow Z$

  1. $x$ and $Y$ are isotopes

  2. $x$ and $Z$ are isobars

  3. $x$ and $Y$ are isobars

  4. $x$ and $Z$ are isotopes


Correct Option: A

From the following the wrong statement is:

  1. Half-life of a free neutron is $10.3$ minutes

  2. The stability of a nucleus is only determined by the number of neutrons present in it.

  3. Both fast and slow neutrons are capable of penetrating the nucleus

  4. A free neutron decays into a proton, an electron and positron


Correct Option: D
Explanation:

A free neutron will decay with a half life of about $10.3$ minutes but it is stable if combined into a nucleus. The stability of a nucleus is determined by number of neutrons as well as protons, Only fast moving neutrons are capable of penetrating the nucleus.A few neutron decays into a proton, an electron and anti-neutrino, and other options are known facts.

When $ _{3}Li^{7}$ nuclei are bombarded by protons, and the resultant nuclei are $ _{4}Be^{8}$ , the emitted particles will be.   

  1. alpha particles

  2. beta particles

  3. gamma photons

  4. neutrons


Correct Option: C
Explanation:

$ _{3}Li^{7} + ^{1} _{1}P \Rightarrow  _{4}Be^{8} + _{0}^{0}\gamma (energy)$
As, the emitted particles is only energy and gamma photons radiation is nothing but energy so, option  C is correct.

$ {27}^{57}\textrm{Co}$ will emit __________ radiation

  1. $\beta^{-}$

  2. $ \beta^{+}$

  3. $\alpha$

  4. electron capture


Correct Option: D
Explanation:

Electron capture :   $^{57} _{27}$  $ + $ $ ^0 _{-1}e$ $\rightarrow$  $^{57} _{26}Fe$

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 nucleus

  3. An isolated neutron can change into a proton

  4. An isolated proton can change int a neutron


Correct Option: B,C
Explanation:

$\beta^{+}$ decay cannot occur in an isolated proton because it requires energy due to the mass of the neutron being greater than the mass of the proton. $\beta^{+}$ decay can only happen inside nuclei when the daughter nucleus has a greater binding energy.


an isolated neutron is not stable outside hence it decays emitting a proton.

If $ _{5}\textrm{B}^{11}$ converts into $ _{6}\textrm{C}^{11}$, then the particle emitted in this process will be 

  1. electron

  2. proton

  3. neutron

  4. positron


Correct Option: A
Explanation:
$\beta ^-$ minus decay produces an electron and electron anti neutrino.
$\beta ^-$ decay increases the proton by $1$ and reduces the neutron number by $1$
Hence, mass number remains same but atomic number increase by $1.$

What does a neutron decays to?

  1. one $p$, one $v$ one $\beta^{+}$

  2. one $\beta{+}$, one$\beta{-}$ and $v$.

  3. one $p$, one $\beta^{-}$ and one $\bar{v}$

  4. all the above


Correct Option: C
Explanation:
The neutron decays into a proton, an electron, and an antineutrino of the electron type.

The particle emitted in the nuclear reaction 
$ _{z}\textrm{X}^{A}$ = $ _{z+1}\textrm{Y}^{A}$ + ..... will be 

  1. $\alpha$ -particle

  2. $\beta^{-}$ -particle

  3. $\beta{+}$ -particle

  4. $Photon$


Correct Option: B
Explanation:
In $\beta^{-}$ decay, there is a gain of one proton and an anti-neutrino is emitted from nucleus, hence atomic number is increased by 1 and weight remains same.

The nucleus of mass  $M +  \Delta m$  is at rest and decays into two daughter nuclei of equal mass  $\dfrac {  M  } { 2 }$  each. Speed of light is  $ c.$ The speed of daughter nuclei is

  1. $c \dfrac { \Delta m } { M + \Delta m }$

  2. $c \sqrt { \dfrac { 2 \Delta m } { M } }$

  3. $c \sqrt { \dfrac { \Delta m } { M } }$

  4. $c \sqrt { \dfrac { \Delta m } { M + \Delta m } }$


Correct Option: B

The particle $X$ in the following nuclear reaction is  $ _{7}^{13}\textrm{N}$ $\longrightarrow $ $ _{6}^{13}\textrm{C}+$  $ _{1}^{0}\textrm{e}$ + $X$

  1. $P$

  2. $v$

  3. $e^{-}$

  4. $\alpha$


Correct Option: B
Explanation:

The given reaction :      $^{13} _7N\rightarrow ^{13} _6C + ^0 _1e + ^A _Z X$

Conservation of mass number :      $13 = 13 + 0 +A$              $\implies A=0$
Conservation of atomic number :      $7 = 6 + 1 +Z$              $\implies Z=0$
Thus, the particle $X$ is neutrino i.e.  $\nu$

A radioactive material initially contains $10gm$ and after few days $3gm$ is left, then the emission rate of $\alpha$ or $\beta$ particle:-

  1. Will continue as usual

  2. Becomes $0.3$times

  3. Increases

  4. None of the above


Correct Option: B

The ionization potential for second He electron is 

  1. 13.6 V

  2. 27.2 V

  3. 54.4 V

  4. 3.4 V


Correct Option: C

Which word equation represents $\beta^+$ decay?

  1. Proton $\rightarrow$ neutron $+$ electron $+$ electron antineutrino

  2. Proton $\rightarrow$ neutron $+$ electron $+$ electron neutrino

  3. Proton $\rightarrow$ neutron $+$ positron $+$ electron antineutrino

  4. Proton $\rightarrow$ neutron $+$ positron $+$ electron neutrino


Correct Option: D
Explanation:

 Positive beta decay $(β+\ \ \text{decay})$ also called the  positron emission. In this decay  a proton in the parent nucleus decays into a neutron that remains in the daughter nucleus, and the nucleus emits a neutrino and a positron, which is a positive particle like an ordinary electron in mass but of opposite charge. Thus, positive beta decay produces a daughter nucleus, the atomic number of which is one less than its parent and the mass number of which is the same.

The equation can be written as :

$p \rightarrow n+ e^+ + v _e$

A radioactive substance contains a number of identical nuclei that emit $\beta$- particles. Which property of these nuclei remains unaltered by emission?

  1. charge

  2. neutron number

  3. nucleon number

  4. proton number


Correct Option: C
Explanation:

We know that in any $\beta -$ emission the mass number of the nucleus does not changes .

This implies that the nucleon number will remain same. 
So, option C is correct.

In $\beta^-$ decay, a

  1. neutron converts into a proton emitting antineutrino.

  2. neutron converts into a proton emitting neutrino.

  3. proton converts into a neutron emitting antineutrino.

  4. proton converts into a neutron emitting neutrino.


Correct Option: A
Explanation:

The reaction of beta decay is shown as :

$n \rightarrow p + e^- + \bar{\nu} _e$               (antineutrino)

So, in a $\beta-$ decay, a neutron converts into a proton emitting antineutrino.

option (A) is correct.

The number of $\beta$-particles, if a radioactive element $ _{90}X^{238}$ decays into $ _{83}Y^{222}$ is :

  1. $4$

  2. $6$

  3. $2$

  4. $1$


Correct Option: D

Which of the following nuclei is produced when a $ _{92}U^{238}$ nucleus undergoes a $(d, 2n)$ reaction followed by a beta decay?

  1. $ _{93}Np^{238}$

  2. $ _{94}Pu^{239}$

  3. $ _{94}Pu^{238}$

  4. $ _{92}U^{238}$


Correct Option: C

In which of the following processes, the number of protons in the nucleus increase?

  1. $\alpha-decay$

  2. $\beta^--decay$

  3. $\beta^+-decay$

  4. k-capture


Correct Option: B
Explanation:

${ \beta  }^{ - }\quad decay$ :
Example: $ _{ 6 }^{ 14 }{ C }\longrightarrow _{ 7 }^{ 14 }{ N }+{ e }^{ - }+\bar { { v } _{ e } }$
Atomic number increases by 1 unit, which implies increase in proton.

Atomic masses of two isobars $ _{29}^{63}Cu$ and $ _{30}^{64}Zn$ are $63.9298 u$ and $63.9292 u$, respectively. It can be concluded from this data that

  1. both the isobars are stable

  2. $^{64}Zn$ is radioactive, decaying to $^{64}Cu$ through $\beta-decay$

  3. $^{64}Cu$ is radioactive, decaying to $^{64}Zn$ through $\beta-decay$

  4. $^{64}Cu$ is radioactive, decaying to $^{64}Zn$ through $\gamma-decay$


Correct Option: C
Explanation:

$Zn$ with higher no.of nucleons has a lower mass than $Cu$, which means that the binding energy/nucleon is higher in $Zn$.
Which means that $Zn$ is more stable than $Cu$.
Hence, $Cu$ will have a tendency to convert to $Zn$ by radioactive decay
$\beta$ decay to change the atomic number.
$\gamma$ decay can't help in changing the no. of protons in the nucleus.

The electron emitted in beta radiation originates from

  1. inner orbits of atoms

  2. free electrons existing in nuclei

  3. decay of a neutron in a nucleus

  4. photon escaping from the nucleus


Correct Option: C
Explanation:

The electron emitted in beta radiation may originates from neutron and it increases the atomic number $1$.

Masses of two isobars $ _{29}Cu^{64}$ and $ _{30}Zn^{64}$ are $63.9298\ u$ and $63.9292\ u$, respectively. It can be conclude from these data that

  1. Both the isobars are stable

  2. $Zn^{64}$ is radioactive, decaying to $Cu^{64}$ through $\beta-decay$

  3. $Cu^{64}$ is radioactive, decaying to $Zn^{64}$ through $\gamma-decay$

  4. $Cu^{64}$ is radioactive, decaying to $Zn^{64}$ through $\beta-decay$


Correct Option: D
Explanation:

Same no. of nucleons for $Cu^{64}$ and $Zn^{64}$
However, $M _{Cu}> M _{Zn}$
which indicates that the mass defect/ nuclear binding energy per nucleon is lesser of $Cu$, hence it will have a tendency to get to more stable form by $\beta $ decay.

Neutron decay in free space is given as follows
$ _{ 0 }{ n }^{ 1 }\longrightarrow _{ 1 }{ H }^{ 1 }+ _{ 1 }{ e }^{ 0 }+$[  ]
Then the parenthesis [  ] represents a

  1. neutrino

  2. photon

  3. antineutrino

  4. graviton


Correct Option: C
Explanation:

Since the charge is already conserved in the decay process, proton cannot be ejected.

To conserve the spin angular momentum of the initial neutron particle, an antineutrino (chargeless) is ejected.

The number of neutrons in the element L in the following nuclear changes is 
$^{238} _{92}M\, \rightarrow\, ^x _y\, N\, +\, ^4 _2\, He$
$^X _YN\, \rightarrow\, ^A _BL\, +\, 2\beta^+$

  1. $146$

  2. $144$

  3. $140$

  4. $142$


Correct Option: A
Explanation:
$ _{ 92 }^{ 238 }{ M }\rightarrow _{ y }^{ x }{ N }+ _{ 2 }^{ 4 }{ He }$
$y=92-2=90 \\ x=238-4=234$
$ _{ 90 }^{ 234 }{ N }\rightarrow _{ B }^{ A }L+2 _{ +1 }^{ o }{ \beta  }\\ B=90-2=88\\ A=234$
no. of neutrons $=234-88 \\ =146$
So, $(A)$ is correct option.

$^{11} _{6}C\, \rightarrow\,  ^{11} _{5}B$ decay produces -

  1. Positron

  2. $\beta$-particle

  3. $\alpha $-particle

  4. None of these


Correct Option: A
Explanation:

The positron or antielectron is the antiparticle or the antimatter counterpart of the electron. The positron has an electric charge of $+1e$, a spin of $\cfrac{1}{2}$ and has the same mass as an electron.

In radioactive decay process, the emitted negatively charged $\beta$ - particles are :

  1. the electrons present inside the nucleus

  2. the electrons produced as a result of the decay of neutrons inside the nucleus

  3. the electrons produced as a result of collisions between atoms

  4. the electrons orbiting around the nucleus


Correct Option: B
Explanation:

Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$ - decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state.
In negative $\beta$- decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted $\beta$- particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

Which of the following statement is correct?

  1. The rest mass of a stable nucleus is less than the sum of the rest masses of ts separated nucleons.

  2. The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons

  3. In nuclear fission, energy is released by fusion two nuclei of medium mass (approximately 100 amu).

  4. In nuclear fission, energy is released by fragmentation of a very low nucleus.


Correct Option: A

Find out the missing particle in the following nuclear reaction?
$^2 _1H+^{63} _{29}Cu \rightarrow ^{64} _{30}Zn+(?)$

  1. Proton

  2. Neutron

  3. Electron

  4. Positron

  5. Deuteron


Correct Option: B
Explanation:

The given reaction :    $^2 _1 H$  $+$  $^{63} _{29} Cu  \rightarrow      ^{64} _{30} Zn  $  $+$  $^A _Z X$

In a nuclear reaction, mass number and the atomic number are conserved.
$\therefore$    $2+ 63  = 64 + A$   $\implies A = 1$
Also,  $1 + 29  = 30+Z$    $\implies Z = 0$
Thus, the missing particle is $^1 _0 n$  i.e  neutron.

The number of neutrons decreases by 1 after radioactive decay. Identify the type of decay.

  1. Alpha decay

  2. ${\beta}^{-}$ decay

  3. ${\beta}^{+}$ decay

  4. Electron capture

  5. Gamma decay


Correct Option: B
Explanation:

We know that for $\beta^-$ decay, one neutron is converted into proton and electron and the electron is ejected in the reaction. For this, the number of neutrons will decrease by one unit. 

But in $\beta^+$  decay , the positron is ejected and neutron number will increase by one unit. 

Compared to the parent nucleus, the daughter nucleus of a $\beta$ decay has:

  1. the same mass number but a smaller atomic number

  2. the same mass number but a greater atomic number

  3. a greater mass number but the same atomic number

  4. a smaller mass number but the same atomic number

  5. None of the above


Correct Option: B
Explanation:

When beta decay occurs, the atomic number of nucleus will increase by one unit and mass number will remain same. Thus, here the after beta decay the daughter nucleus has same mass number as product nuclei and one unit atomic number will increase than parent nuclei.    

When carbon $-14$ undergoes beta (electron) decay, it transmutes into what?

  1. $Carbon-13$

  2. $Carbon-12$

  3. $Nitrogen-13$

  4. $Nitrogen-12$

  5. $Nitrogen-14$


Correct Option: E
Explanation:
Beta decay of Carbon-14 : $^{14} _{6}   $ $ \rightarrow$  $^{A} _ZX $ $+$  $^{0} _{-1} e$

Conservation of mass number :  $14 = A+ 0$    $\implies A = 14$
Conservation of atomic number :  $6 = Z-1$    $\implies Z =7$
Thus the daughter nucleus is: $^{14} _{7} N$  or  $Nitrogen- 14$.

Find out the product of a $Co^{60}$ atom that undergoes one beta plus decays?

  1. iron-60

  2. manganese-60

  3. copper-60

  4. copper-62

  5. iron-62


Correct Option: A
Explanation:

$\beta^+$ decay: $^{60} _{28}Co \rightarrow$ $^{A} _{Z}X$ $+$ $^0 _{+1}e$ 

In a nuclear reaction, the mass number as well as the atomic number is conserved.
$\therefore$  $60 = 0 + A$ $\implies A =60$
Also, $27 = 1 + Z$  $\implies Z = 26$
Thus, the product of the reaction is iron-60  i.e $^{60} _{26}Fe$.

In $\beta$ decay.

  1. Atomic number decreased by one

  2. Mass number decreases by one

  3. Proton number remains the same

  4. Neutron number decreases by one


Correct Option: D
Explanation:

Beta decay of $ _Z^A X$ :    $ _Z^A X \rightarrow   _{Z+1}^A Y$ $+$ $  ^o _{-1}e$

Number of protons in X: $p _x = Z $
Number of neutrons in X: $n _x = A-Z$

Number of protons in Y: $ p _y= Z+1 $
Number of neutrons in X: $ n _y= A-(Z+1) =  A-Z - 1 $

$\implies$  $n _y = n _x -1$
Thus neutron number decreases by one in beta decay.

When an atom undergoes $\beta$-decay, its atomic number

  1. Does not change

  2. Increases by $1$

  3. Decreases by $1$

  4. Increases by $2$


Correct Option: B
Explanation:
When an atom undergoes $\beta$-decay the atomic number increases by 1.

When an atom undergoes $\beta$-decay, one of the neutrons breaks into one proton and one electron. The resultant electron is then ejected out of the nucleus and this is called as the $\beta$ particle. While the resultant proton stays inside the nucleus which results in increase of atomic number by 1, whereas the atomic mass remains invariant.

Hence the correct answer is option B.

10 grams of $^{57}Co$ kept in an open container beta-decays with a half-life of $270$ days. The weight of the material inside the container after $540$ days will be very nearly.

  1. $10g$

  2. $5g$

  3. $25g$

  4. $125g$


Correct Option: A
Explanation:

$^{57}Co$ is undergoing beta decay i.e electron is being produced.But an electron has very less mass ($9.11\times10^{-31}kg$) as compared to the $Co$ atom.therefore,after $570$ days,even the atom go under large beta decay,the weight of the material in the container will be nearly $10 g$.

During a $\beta ^-$ decay which of the following statements are correct?

  1. The daughter nucleus has one proton less than the parent nucleus

  2. The daughter nucleus has one proton more than the parent nucleus

  3. An electron which is already present within the nucleus is rejected

  4. A neutron in the nucleus decays emitting an electron


Correct Option: B
Explanation:
$\beta ^- Decay$
$^\theta _{-1}e$ or $^o _{-1}\beta \rightarrow$ Symbols for Beta particles emitted during $\beta^-$ decay.
$\theta \rightarrow$ electron is emitted
Eg- $ _{ 53 }^{ 131 }{ I }\rightarrow _{ 54 }^{ 131 }{ Xe }+ _{ -1 }^{ 0 }{ \beta  }+\bar { { \upsilon  } _{ e } } $
where,  $ _{ -1 }^{ 0 }{ \beta  }$ is $\beta$ particle
$\bar { { \upsilon  } _{ e } }$ is antineutrino
So, $(B)$ is correct option.

Masses of neutron, proton and electron are $1.0087$U, $1.0073$u and $0.0005$u respectively. If a neutron decays into a proton and an electron, the energy released would be about.

  1. $0.68$ MeV

  2. $0.84$ MeV

  3. $0.75$ MeV

  4. $1.22$ MeV


Correct Option: B
Explanation:

$^1 _0n\rightarrow$ $^1 _1p+$ $^0 _{-1}e$
$Q=(m _n=m _p-m _e)c^2$

$=0.0009\times 931.5$.
0.84

During $\beta-decay$ (beta minus), the emission of antineutrino particle is supported by which of the following statement(s)?

  1. Angular momentum conservation holds good in any nuclear reaction.

  2. Linear momentum conservation holds good in any nuclear reaction.

  3. The KE of emitted $\beta-particle$ is varying continuously to a maximum value.

  4. None of the above.


Correct Option: A,B,C
Explanation:

The following principles hold true in any situation in the universe
1. Conservation of linear momentum (if no external force)
2. Conservation of angular momentum (no external moment)
2. Conservation of energy
3. Conservation of charge

A positron is emitted from $\mathrm{N}\mathrm{a} _{11}^{23}$. The ratio of the atomic mass and atomic number of the resulting nuclide is

  1. $22/10$

  2. $22/11$

  3. $23/10$

  4. $23/12$


Correct Option: C
Explanation:

When a positron is emitted from nucleus, proton will convert into a neutron. So atomic number will be decrease by one  but atomic mass will be constant. 

Thus, atomic mass / atomic number $=23/(11-1)=23/10 $

$ _{6}^{11}\textrm{C}$ on decay produces 

  1. positron

  2. $\alpha-$ particle

  3. $\beta-$ particle

  4. $ _{5}^{11}\textrm{B}$


Correct Option: A,C,D
Explanation:

$ _{6}^{11}\textrm{C}\rightarrow _{5}^{11}\textrm{B}+ _{+1}^{0}\textrm{e}$

Masses of two isobars $ _{29}^{64}\textrm{Cu}$ and $ _{30}^{64}\textrm{Zn}$ are $63.9298 amu$ and $63.9292 amu$ respectively. It can be concluded from these data that 

  1. Both the isobars are stable

  2. $^{64}Zn$ is radioactive, decaying to $^{64}Cu$ through $\beta -$ decay

  3. $^{64}Cu$ is radioactive, decaying to $^{64}Zn$ through $\lambda -$ decay

  4. $^{64}Cu$ is radioactive, decaying to $^{64}Zn$ through $\beta -$ decay


Correct Option: D
Explanation:

Since mass of the isobar has decreased, an electron has been ejected.

Thus $^{64} _{29}Cu$ decays to $^{64} _{30}Zn$ through the $\beta - $ decay as-
$^{64} _{29}Cu\rightarrow^{64} _{30}Zn+^{0} _{-1}e$

A nucleus of magnesium decays into a nucleus of sodium by emitting a $\beta^{+}$ particle. The decay is
represented by the equation shown.
$^{23} _{12}Mg \rightarrow ^{P} _{Q}Na + ^{0} _{+1}\beta$
What are the values of $P$ and $Q$?

  1. $P = 22, Q = 11$

  2. $P = 22, Q = 13$

  3. $P = 23, Q = 11$

  4. $P = 23, Q = 13$


Correct Option: C
Explanation:

Since the no. of nucleons will conserved . 

So, comparing the no. of proton on both sides
$12 = Q+1$
$\implies Q = 11$
By conservation of mass 
$23 = P+0$
$\implies P = 23$

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