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

The equation $ _{88}Ra^{226}\rightarrow _{86}Rn^{222}+ _{2}He^{4}$ emits which particle?

  1. $\beta$-decay
  2. $\alpha$-decay
  3. $\gamma$-decay

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\alpha $ - decay 

When an unstable atomic nucleus emits two protons and two neutrons the radioactive process is known as alpha decay. 

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

During $\beta^-$ emission:

  1. a neutron in the nucleus decays emitting an electron

  2. an atomic electron is ejected

  3. an electron already present within the nucleus is ejected

  4. a part of the binding energy of the nucleus is converted into an electron

  5. a proton in the nucleus decays emitting an electron
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\beta^-$ emission is due to decay of neutron in the nucleus $n\rightarrow p+e^-$.

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

Nuclei of a radioactive element $A$ are being produced at a constant rate $\alpha$. The element has a decay constant $\lambda$. At $t =0$, there are $N _{0}$ nuclei of the element.
If $\alpha = 2N _{0}\lambda$, calculate the number of nuclei of $A$ after one half life of $A$, and also the limiting value of $N$ as $t\rightarrow \infty$.

  1. $\dfrac {4N _{0}}{2}, 2N _{0}$.
  2. $\dfrac {3N _{0}}{2}, 2N _{0}$.
  3. $\dfrac {5N _{0}}{2}, 2N _{0}$.
  4. $\dfrac {6N _{0}}{2}, 2N _{0}$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The rate equation is dN/dt = alpha - lambda*N. Solving this with N(0)=N0 and alpha=2*N0*lambda leads to N(t) = 2*N0 - N0*exp(-lambda*t). At t = half-life (ln2/lambda), N = 2*N0 - N0/2 = 1.5*N0. As t approaches infinity, N approaches 2*N0.

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

$90$% of a radioactive sample is left undecayed after time $t$ has elapsed. What percentage of the intial sample will decay in a total time $2t$:

  1. $20$%
  2. $19$%
  3. $40$%
  4. $38$%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

After time t, 90% remains (N/N0 = 0.9). After time 2t, the fraction remaining is (0.9)^2 = 0.81. The amount decayed is 1 - 0.81 = 0.19, or 19%.

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

$ _{84}P _{0}^{210}$ originally at rest emits $\alpha $- particles of KE 'K' Find  the KE of recoiling nucleus:

  1. $\dfrac{4}{214}K$
  2. $\dfrac{4}{206}K$
  3. $\dfrac{K}{206}$
  4. $\dfrac{K}{214}K$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$84P _o^{210}\xrightarrow 82Pb^{206}+\alpha \Rightarrow $ mass of remaining nucleus $(Pb)=206$

Initial momentum =$0$ (initial Po was at rest)
Momentum of $\alpha$- particle carrying $KE=K$
$P _1\sqrt{2m _{\alpha}k}$     $m _{alpha}=4n$,mass of $\alpha$
$P _1=\sqrt{8k}$
from conservation of momentum
$P _1+P _2=0$
$\Rightarrow P _1^2=P _2^2$
$2\ mrk'=2k$
$K'=\dfrac{4k}{mr}=\dfrac{4k}{206}$

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

When a radioactive nucleus emits a $\beta $- particular, the proton- neutron ratio

  1. decreases

  2. increases

  3. remains same

  4. first decreases and increases
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When the neutron to proton ratio in the nucleus is too great a beta particle is emitted. In basic beta decay a neutron is transformed into a proton and an electron. The electron is then emitted as a beta particle which increases the atomic number by one and the molar mass is unchanged.

Hence, the proton- neutron ratio increase.

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

A free neutron is unstable against $\beta$ decay with a half life of about $600$ seconds:

  1. The expression of this decay process in $n\rightarrow p+e^{-}+\vec{v}$
  2. If three are $600$ free neutrons initially, the time by which $450$ of them have decayed is $2400$ sec.
  3. The dacay rate of the sample is $0.593$ Bq.
  4. The dacay rate of the sample is $593$ Bq.
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Initial number of nuclei of a radioactive substance is $5 \times 10 ^ { 16 }$ and half-life is $10$ yrs. Find the number of nuclei decayed in $5$ yrs.

  1. $2 \times 10 ^ { 16 }$
  2. $1.5 \times 10 ^ { 16 }$
  3. $3.5 \times 10 ^ { 16 }$
  4. $2.5 \times 10 ^ { 16 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Remaining$ nuclei after $5$ years will be $N=5\times 10^{16} \times (\dfrac{1}{2})^{5/10}=\dfrac{5\times 10^{16}}{\sqrt[2]{2}}=\dfrac{5\times 10^{16}}{1.414}=3.54\times 10^{16}$


So the decayed nuclei will be $(5-3.54)\times 10^{16}=1.46\times 10^{16}$
nearly $1.5\times 10^{16}$

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

A mixture consists of two radioactive materials ${ A } _{ 1 }$ and ${ A } _{ 2 }$ with half lives of 20 s and 10 s respectively. Initially the mixture has $40 g$ of ${ A } _{ 1 }$ and $160 g$ of ${ A } _{ 2 }$. The active amount of the two in the mixture will become equal after :

  1. $20s$
  2. $40s$
  3. $60s$
  4. $80s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

N1(t) = 40 * (1/2)^(t/20). N2(t) = 160 * (1/2)^(t/10). Set N1(t) = N2(t): 40 * (1/2)^(t/20) = 160 * (1/2)^(t/10). Dividing by 40: (1/2)^(t/20) = 4 * (1/2)^(t/10). This simplifies to (1/2)^(t/20) = 2^2 * (1/2)^(t/10). Solving for t gives 40s.

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

Samples of two radioactive nuclides $A$ and $B$ are taken. $\lambda _ { A }$ and $\lambda _ { B }$ are the disintegration constants of $A$ and $B$ respectively. In which of the following cases, the two samples can simultaneously have the same decay rate at any time ? 

  1. Initial rate of decay of $A$ is twice the initial rate of decay of $B$ and $\lambda _ { A } = \lambda _ { B }$
  2. Initial rate of decay of $A$ is twice the initial rate of decay of $B$ and $\lambda _ { A } > \lambda _ { B }$
  3. Initial rate of decay of $B$ is twice the initial rate of decay of $A$ and $\lambda _ { A } > \lambda _ { B }$
  4. Initial rate of decay of $B$ is same as the rate of decay of $A$ at t = 2h and $\lambda _ { B } < \lambda _ { A }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} N={ N _{ 0 } }{ e^{ -\lambda t } } \ \therefore if\, initial\, rate\, is\, same\, \, and\, { \lambda _{ A } }={ \lambda _{ B } } \end{array}$

Hence,
option $(A)$ is correct answer.