Tag: change in nucleus due to radioactive decay

Questions Related to change in nucleus due to radioactive decay

During a $\beta ^-$ decay which of the following statements are correct?

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

  1. The daughter nucleus has one proton less than the parent nucleus

  2. The daughter nucleus has one proton more than the parent nucleus

  3. An electron which is already present within the nucleus is rejected

  4. A neutron in the nucleus decays emitting an electron

Show answer
Correct Option: B
Explanation:
$\beta ^- Decay$
$^\theta _{-1}e$ or $^o _{-1}\beta \rightarrow$ Symbols for Beta particles emitted during $\beta^-$ decay.
$\theta \rightarrow$ electron is emitted
Eg- $ _{ 53 }^{ 131 }{ I }\rightarrow _{ 54 }^{ 131 }{ Xe }+ _{ -1 }^{ 0 }{ \beta  }+\bar { { \upsilon  } _{ e } } $
where,  $ _{ -1 }^{ 0 }{ \beta  }$ is $\beta$ particle
$\bar { { \upsilon  } _{ e } }$ is antineutrino
So, $(B)$ is correct option.

Masses of neutron, proton and electron are $1.0087$U, $1.0073$u and $0.0005$u respectively. If a neutron decays into a proton and an electron, the energy released would be about.

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

  1. $0.68$ MeV

  2. $0.84$ MeV

  3. $0.75$ MeV

  4. $1.22$ MeV

Show answer
Correct Option: B
Explanation:

$^1 _0n\rightarrow$ $^1 _1p+$ $^0 _{-1}e$
$Q=(m _n=m _p-m _e)c^2$

$=0.0009\times 931.5$.
0.84

During $\beta-decay$ (beta minus), the emission of antineutrino particle is supported by which of the following statement(s)?

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

  1. Angular momentum conservation holds good in any nuclear reaction.

  2. Linear momentum conservation holds good in any nuclear reaction.

  3. The KE of emitted $\beta-particle$ is varying continuously to a maximum value.

  4. None of the above.

Show answer
Correct Option: A,B,C
Explanation:

The following principles hold true in any situation in the universe
1. Conservation of linear momentum (if no external force)
2. Conservation of angular momentum (no external moment)
2. Conservation of energy
3. Conservation of charge

A positron is emitted from $\mathrm{N}\mathrm{a} _{11}^{23}$. The ratio of the atomic mass and atomic number of the resulting nuclide is

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

  1. $22/10$

  2. $22/11$

  3. $23/10$

  4. $23/12$

Show answer
Correct Option: C
Explanation:

When a positron is emitted from nucleus, proton will convert into a neutron. So atomic number will be decrease by one  but atomic mass will be constant. 

Thus, atomic mass / atomic number $=23/(11-1)=23/10 $

$ _{6}^{11}\textrm{C}$ on 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. $\alpha-$ particle

  3. $\beta-$ particle

  4. $ _{5}^{11}\textrm{B}$

Show answer
Correct Option: A,C,D
Explanation:

$ _{6}^{11}\textrm{C}\rightarrow _{5}^{11}\textrm{B}+ _{+1}^{0}\textrm{e}$

Masses of two isobars $ _{29}^{64}\textrm{Cu}$ and $ _{30}^{64}\textrm{Zn}$ are $63.9298 amu$ and $63.9292 amu$ respectively. It can be concluded 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. $^{64}Zn$ is radioactive, decaying to $^{64}Cu$ through $\beta -$ decay

  3. $^{64}Cu$ is radioactive, decaying to $^{64}Zn$ through $\lambda -$ decay

  4. $^{64}Cu$ is radioactive, decaying to $^{64}Zn$ through $\beta -$ decay

Show answer
Correct Option: D
Explanation:

Since mass of the isobar has decreased, an electron has been ejected.

Thus $^{64} _{29}Cu$ decays to $^{64} _{30}Zn$ through the $\beta - $ decay as-
$^{64} _{29}Cu\rightarrow^{64} _{30}Zn+^{0} _{-1}e$

A nucleus of magnesium decays into a nucleus of sodium by emitting a $\beta^{+}$ particle. The decay is
represented by the equation shown.
$^{23} _{12}Mg \rightarrow ^{P} _{Q}Na + ^{0} _{+1}\beta$
What are the values of $P$ and $Q$?

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

  1. $P = 22, Q = 11$

  2. $P = 22, Q = 13$

  3. $P = 23, Q = 11$

  4. $P = 23, Q = 13$

Show answer
Correct Option: C
Explanation:

Since the no. of nucleons will conserved . 

So, comparing the no. of proton on both sides
$12 = Q+1$
$\implies Q = 11$
By conservation of mass 
$23 = P+0$
$\implies P = 23$