Tag: change in nucleus due to radioactive decay

Questions Related to change in nucleus due to radioactive decay

In $\beta^-$ decay, a

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

  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.

Show answer
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 :

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

  1. $4$

  2. $6$

  3. $2$

  4. $1$

Show answer
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?

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

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

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

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

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

Show answer
Correct Option: C

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

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

  1. $\alpha-decay$

  2. $\beta^--decay$

  3. $\beta^+-decay$

  4. k-capture

Show answer
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

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

  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 $\gamma-decay$

Show answer
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

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

  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

Show answer
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

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

  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 $\beta-decay$

Show answer
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

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

  1. neutrino

  2. photon

  3. antineutrino

  4. graviton

Show answer
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 
$^{238} _{92}M\, \rightarrow\, ^x _y\, N\, +\, ^4 _2\, He$
$^X _YN\, \rightarrow\, ^A _BL\, +\, 2\beta^+$

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

  1. $146$

  2. $144$

  3. $140$

  4. $142$

Show answer
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 -

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

  1. Positron

  2. $\beta$-particle

  3. $\alpha $-particle

  4. None of these

Show answer
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.