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Gamma decay - class-XI

Description: gamma decay
Number of Questions: 21
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Tags: physics atomic nuclei nuclei atomic, nuclear and particle physics
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A free nucleus of mass $24$ u emits a gamma photon [when initially at rest]. The energy of the photon is $7$ MeV. The recoil energy of the nucleus in keV is 

  1. $1.1$

  2. $1.2$

  3. $1.0$

  4. $1.3$


Correct Option: A

Hydrogen atom will be in its ground state,if its electron is in

  1. any energy level

  2. the lowest energy state

  3. the highest energy state

  4. the intermediate state


Correct Option: B
Explanation:

Hydrogen atom has one electron. Depending upon the electron residing in which energy state hydrogen energy varies. So, for hydrogen atom to be in ground state electron should be in lowest energy state.

When writing electron configurations,electrons are represented in their lowest possible energy state.Which of the following state is this ?

  1. excited state.

  2. lowest state.

  3. configuration state.

  4. ground state.

  5. unenergetic state.


Correct Option: D
Explanation:

According to Aufbau's principle, electrons are filled in the atomic orbitals in order of lower energy to the higher energy starting from the orbital of lowest possible energy known as the ground state.

A beam of ultraviolet radiation having wavelength between $100nm$ and $200nm$ is incident on a sample of atomic hydrogen gas. Assuming that the atoms are in ground state, which wavelengths will have low intensity in the transmitted beam? 

  1. $104nm$

  2. $103nm$

  3. $105nm$

  4. $100nm$


Correct Option: B
Explanation:

Energy corresponding to wavelength 100nm, $\dfrac{1242eV}{100}=12.42 eV$

Energy corresponding to wavelength 200nm, $\dfrac{1242eV}{200}=6.21 eV$
Energy required from ground state to first excited stage:
$E _2-E _1=13.6-3.4=10.2 eV$
Energy required from ground state to second excited stage:
$E _3-E _1=13.6-13.6-1.5=12.1  eV$

Energy required from ground state to third excited stage:
$E _3-E _1=13.6-0.85=12.75  eV$

At 10.2 eV,
$Wavelength 1=\dfrac{1242}{10.2}=122nm$

At 12.1 eV,
$Wavelength 2=\dfrac{1242}{12.1}=103nm$



The ground state energy of the electron in hydrogen atom is equal to :

  1. the ground state energy of the electron in $ { He }^{ + } $

  2. the first excited state energy of the electron in $ { He }^{ + } $

  3. the first excited state energy of the electron in $ { Li }^{ +2 } $

  4. the ground state energy of the electron in $ { Be }^{ +3 } $


Correct Option: B

The total energy of an electron in the ground state of hydrogen atom is $-13.6\space eV$. The potential energy of an electron in the ground state of $Li^{2+}$ ion will be

  1. $122.4\space eV$

  2. $-122.4\space eV$

  3. $244.8\space eV$

  4. $-244.8\space eV$


Correct Option: D
Explanation:

Therefore, for ${ Li }^{ 2+ }$ ion, Total energy is $- 13.6\times 9 = 122.4\ eV$
But, $-K.E= T.E= \dfrac { P.E }{ 2 } $
Therefore, $P.E$ is $-244.8\ eV$

The ground state energy of Hydrogen atom is $–13.6eV$. The potential energy of the electron in this state is:

  1. $0eV$

  2. $-27.2eV$

  3. $1eV$

  4. $2eV$


Correct Option: B
Explanation:

The ground state energy of hydrogen atom $=13.6eV$

Potential energy $=2$ energy of electron
                             $=2(-13.6\ eV)$
                             $=-27.2\ eV$

In which of the following systems will the radius of the first orbit (n = 1) be minimum ?

  1. hydrogen atom

  2. deuterium atom

  3. singly ionized helium

  4. doubly ionized lithium.


Correct Option: D
Explanation:

Radius of the first orbit of an atom  $R _1 = \dfrac{0.529}{Z}$  $A^o$
$\implies \ R _1 \propto\dfrac{1}{Z}$
where $Z$ is the atomic number of Hydrogen-like atom.
Since $Z$ is maximum for doubly ionized lithium, thus radius of first orbit is minimum in doubly ionized lithium.

What is energy released in the $\beta  - decay\;of{\;^{32}}P{ \to ^{32}}S?$(Given:atomic masses:31.97391 u for $\left( {^{32}P\;and\;31.97207\;u\;fo{r^{32}}S} \right)$

  1. -1.2 MeV

    • 1.7 MeV
  2. +2.1 MeV

  3. -0.9 Mev


Correct Option: B
Explanation:

Released energy will be corresponding the $mass $ $defect$ $\text{which is difference in mass of parent nuclei and daughter nuclei}$

so mass defect is $31.97391u- 31.97207 u=0.00184u=0.00184\times 931Mev=1.7Mev$ 
as $1u=931MeV $ of $ energy$.
Option B is correct.

A sample originally contained $10 ^ { 20 }$radioactive atoms, which emit $\alpha$ -particles emitted inthe thirdyear to that emitted during the second year is $0.3 $.How many $\alpha$ particles were emitted in the first year?

  1. $7 \times 10 ^ { 19 }$

  2. $3 \times 10 ^ { 19 }$

  3. $5 \times 10 ^ { 18 }$

  4. $3 \times 10 ^ { 18 }$


Correct Option: A

Nuclei $X$ decay into nuclei $Y$ by emitting $\alpha$ particles. Energies of $\alpha$ particle are found to be only $1MeV$ & $1.4MeV$. Disregarding the recoil of nuclei $Y$. The energy of $\gamma$ photon emitted will be:

  1. $0.8MeV$

  2. $1.4MeV$

  3. $1 MeV$

  4. $0.4MeV$


Correct Option: C

A $ _6C^{12} $ nucleus is to be divided into 3 alpha particles . the amount of energy required to achieve this ( mass of an alpha particle=4.00388 u ) is

  1. 3.405 MeV

  2. 10.837 MeV

  3. 8.133 MeV

  4. 12.573 MeV


Correct Option: B

A source of energy of 100 W is producing energy by fission of 1 Kg $U^{235}$. How long it can kept generation of energy :- 

  1. $2.5 \times 10^4 yr$

  2. $10^6 s$

  3. $8.6 \times 10^7 s$

  4. $100 yr$


Correct Option: A

The atomic masses of the hydrogen isotopes are
$m _1H^1=1.007825\ amu$
$m _1H^2=2.014102\ amu$
$m _1H^3=3.016049\ amu$
The energy released in the reaction
$ _1H^2+ _1H^2 \rightarrow _1H^3+ _1H^1$
is nearly:

  1. $1\ MeV$

  2. $2\ MeV$

  3. $4\ MeV$

  4. $8\ MeV$


Correct Option: C
Explanation:
Mass defect: $\Delta M = 2m _1H^2 -(m _1H^1+m _1H^3)$
$\therefore$   $\Delta M = 2(2.014202)  -(1.007825+3.016049) = 4.33\times 10^{-3}$  amu
Energy released, $E = \Delta M\times 931$  $MeV$         
$\implies$   $E = 4.33\times 10^{-3}\times 931 \approx 4MeV$

A gamma ray photon creates an electron-positron pair. If the rest mass energy of an electron is $0.5MeV$ and the total kinetic energy of the electron-positron pair is $0.78 MeV$, then the energy of the gamma ray photon must be

  1. $0.78MeV$

  2. $1.78MeV$

  3. $1.28MeV$

  4. $0.28MeV$


Correct Option: B
Explanation:

Energy of $\gamma $-rays photon $=$ Rest mass energy $+$ $K.E$ 

                                         $=2\left( 0.5 \right) +0.78\ = 1.78\ MeV$

A stationary nucleus of mass $24\ amu$ emits a gamma photon. The energy of the emitted photon is $7\ MeV$. The recoil energy of the nucleus is:

  1. $2.2\ keV$

  2. $1.1 keV$

  3. $3.1\ keV$

  4. $22\ keV$


Correct Option: B
Explanation:

The energy of emitted photon, $E=hf=7 MeV$

If $p$ be the momentum of photon and $v$ be the recoil velocity of nucleus, then by conservation of momentum 
$p=mv$ or $E/c=mv$ or $v=E/mc$
Thus, recoil energy $K=\dfrac{1}{2}mv^2=\dfrac{m}{2}\times \dfrac{E^2}{m^2c^2}=\dfrac{E^2}{2mc^2}=\dfrac{(7 MeV)^2}{2\times (24\times 931.5 MeV)}=1.09 \times 10^{-3} MeV=1.1 keV$
where $(1 amu=931.5 meV)$

Consider the following nuclear reaction:
$X^{200}\rightarrow A^{110}+B^{90}+Energy$
If the binding energy per nucleon for $X$, $A$ and $B$ are $7.4\ MeV$, $8.2\ MeV$ and $8.2\ MeV$ respectively, the energy released will be:

  1. $90\ MeV$

  2. $110\ MeV$

  3. $200\ MeV$

  4. $160\ MeV$


Correct Option: D
Explanation:

Binding energy of $X$: $E _X = 200\times 7.4 = 1480MeV$

Binding energy of  $A$: $E _A = 110\times 8.2 = 902MeV$
Binding energy of  $B$: $E _B = 90\times 8.2 = 738MeV$
$\therefore$ energy released, $E = E _A+E _B- E _X = 902+738-1480 =160MeV$

In a $\gamma -$decay process, $\gamma-$rays of energy $E$ is emitted. Find the decrease in internal energy of mass $M$ (of nucleus).

  1. $\dfrac{E^2}{2Mc^2}$

  2. $E - \dfrac{E^2}{2Mc^2}$

  3. $E+\dfrac{E^2}{2Mc^2}$

  4. $E+\dfrac{E^2}{Mc^2}$


Correct Option: C
Explanation:

By momentum conservation:
$\cfrac{E}{c} = Mv$
$ v = \cfrac{E}{Mc}$

Now, total decrease in internal energy $=$ Energy of $\gamma$ $+$ $KE$ of $M$
                                                                 $ = E + \cfrac{1}{2} Mv^2$
                                                                 $ = E + \cfrac{E^2}{2Mc^2}$

Mark out the correct statement(s)

  1. in alpha decay, the energy released is shared between alpha particle and daughter nucleus in the form of kinetic energy and share of alpha particle is more than that of the daughter nucleus

  2. in beta decay, the energy released is in the form of kinetic energy of beta particles

  3. in beta minus decay, the energy released is shared between electron and antineutrino

  4. in gamma decay, the energy released is in the form of energy carried by photons termed as gamma rays


Correct Option: A,C,D
Explanation:

in alpha decay, the energy released is shared between alpha particle and daughter nucleus in the form of kinetic energy and share of alpha particle is more than that of the daughter nucleus

The following principles should be appied:
1. Conservation of momentum
2. Conservation of energy
3. Conservation of charge

If alpha has a lower mass
$E = \cfrac{p^2}{2m}$
momentum has to be same for both the product particles, hence lower mass has higher kinetic energy.

in beta minus decay, the energy released is shared between electron and antineutrino In nuclear physics, beta decay (-decay) is a type of radioactive decay in which a proton is transformed into a neutron, or vice versa, inside an atomic nucleus. This process allows the atom to move closer to the optimal ratio of protons and neutrons.
3. in gamma decay, the energy released is in the form of energy carried by photons termed as gamma rays .
released when electrons transit from a higher energy state to a lower energy state,

The wavelength of emitted $\gamma $ rays are in the other

  1. ${ \lambda } _{ \gamma 2 }\quad >\quad { \lambda } _{ \gamma 3 }\quad >{ \quad \lambda } _{ \gamma 1 }$

  2. ${ \lambda } _{ \gamma 3 }\quad >\quad { \lambda } _{ \gamma 2 }\quad >{ \quad \lambda } _{ \gamma 1 }$

  3. ${ \lambda } _{ \gamma 1 }\quad >\quad { \lambda } _{ \gamma 2 }\quad >{ \quad \lambda } _{ \gamma 3 }$

  4. ${ \lambda } _{ \gamma 3 }\quad >\quad { \lambda } _{ \gamma 1 }\quad >{ \quad \lambda } _{ \gamma 2 }$


Correct Option: A
Explanation:

As the frequencies of $\gamma-rays$ are in the order:

$\nu _{\gamma 1} > \nu _{\gamma 3} > \nu _{\gamma 2}$
Thus as wavelength is inversely proportional to frequency.
$\lambda _{\gamma 2} > \lambda _{\gamma 3} > \lambda _{\gamma 1}$
Hence option A is correct

A free nucleus of mass $24 amu$ emits a gamma photon (when initially at rest). The energy of the photon is $7 MeV$. The recoil energy of the nucleus in $keV$ is

  1. $2.2$

  2. $1.1$

  3. $3.1$

  4. $22$


Correct Option: B
Explanation:

$ E = \cfrac{p^2}{2m}$

Conservation of momentum for photon:
$E = \cfrac{hc}{\lambda} = 7 MeV$
$p = \cfrac{h}{\lambda} = 7/c MeV$

Equating the momentum:
$ \cfrac{7}{c }= \sqrt{2E _{nucleus}m}$

Substitute $m = 24\ amu$,
Solving with appropriate units:
$ E _{nucleus} = 1.1\  KeV$

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