Tag: change in nucleus due to radioactive decay

Questions Related to change in nucleus due to radioactive decay

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

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

  1. $2.2$

  2. $1.1$

  3. $3.1$

  4. $22$

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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$

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

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

  1. 1

  2. 2

  3. 4

  4. 6

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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 ................

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

  1. 4n

  2. 4n+3

  3. 4n+2

  4. 4n+1

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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: 

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

  1. $50$

  2. $250$

  3. $29$

  4. $100$

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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:

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

  1. is decreased

  2. is increased

  3. remains the same

  4. first (A) then (B)

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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

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

  1. $1:2$

  2. $1:4$

  3. $1:5$

  4. $1:10$

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Correct Option: A

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

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

  1. $91$

  2. $90$

  3. $89$

  4. $88$

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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 ?

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

  1. electron

  2. $\alpha$-particle

  3. neutron

  4. positron

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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

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

  1. $10.8MeV$

  2. $10.9MeV$

  3. $11MeV$

  4. $11.1MeV$

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Correct Option: A

The antiparticle of electron is

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

  1. positron

  2. $\alpha $-particle

  3. proton

  4. $\beta $-particle

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Correct Option: A
Explanation:

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