Tag: change in nucleus due to radioactive decay

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

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. $22$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Radioactive series are classified by their mass number modulo 4. For a mass number 223, 223 / 4 = 55 with a remainder of 3. Thus, it belongs to the 4n+3 series (Actinium series).

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

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. $100$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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)
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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. $1:10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics nuclei beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

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. $88$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $89$.

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

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

  1. electron

  2. $\alpha$-particle

  3. neutron

  4. positron
Reveal answer Fill a bubble to check yourself
D Correct answer
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}$.