Tag: change in nucleus due to radioactive decay

Questions Related to change in nucleus due to radioactive decay

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:

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

  1. $0.8MeV$

  2. $1.4MeV$

  3. $1 MeV$

  4. $0.4MeV$

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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

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

  1. 3.405 MeV

  2. 10.837 MeV

  3. 8.133 MeV

  4. 12.573 MeV

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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 :- 

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

  1. $2.5 \times 10^4 yr$

  2. $10^6 s$

  3. $8.6 \times 10^7 s$

  4. $100 yr$

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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:

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

  1. $1\ MeV$

  2. $2\ MeV$

  3. $4\ MeV$

  4. $8\ MeV$

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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

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

  1. $0.78MeV$

  2. $1.78MeV$

  3. $1.28MeV$

  4. $0.28MeV$

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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:

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

  1. $2.2\ keV$

  2. $1.1 keV$

  3. $3.1\ keV$

  4. $22\ keV$

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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:

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

  1. $90\ MeV$

  2. $110\ MeV$

  3. $200\ MeV$

  4. $160\ MeV$

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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).

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

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

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

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

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

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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)

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

  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

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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

gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

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

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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