Tag: change in nucleus due to radioactive decay

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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

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.

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

The radius of spherical nucleus as measured by electron scattering is 36. fm. what is the likely mass number of the nucleus?

  1. 27

  2. 40

  3. 56

  4. 120
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Nuclear radius R = R0 * A^(1/3), where R0 is approx 1.2 fm. 36 = 1.2 * A^(1/3). A^(1/3) = 30. A = 30^3 = 27000. This calculation suggests the value 36 fm is very large. If R0 = 1.2, A = 27 is R = 1.2 * 3 = 3.6 fm. The question likely meant 3.6 fm.

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

A bone containing 200 g carbon-14 has a $\beta $ decay rate of 375 deacy/min. Calculate the time that has elapsed since the death of the living one. Given the rate of decay for the living organism is equal to 15 decay per min per gram of carbon and half - life of carbon -14 is 5730 years,

  1. 27190 years

  2. 1190 years

  3. 17190 years

  4. None of these
Reveal answer Fill a bubble to check yourself
A 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

From the following the wrong statement is:

  1. Half-life of a free neutron is $10.3$ minutes

  2. The stability of a nucleus is only determined by the number of neutrons present in it.

  3. Both fast and slow neutrons are capable of penetrating the nucleus

  4. A free neutron decays into a proton, an electron and positron
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A free neutron will decay with a half life of about $10.3$ minutes but it is stable if combined into a nucleus. The stability of a nucleus is determined by number of neutrons as well as protons, Only fast moving neutrons are capable of penetrating the nucleus.A few neutron decays into a proton, an electron and anti-neutrino, and other options are known facts.

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

When $ _{3}Li^{7}$ nuclei are bombarded by protons, and the resultant nuclei are $ _{4}Be^{8}$ , the emitted particles will be.   

  1. alpha particles

  2. beta particles

  3. gamma photons

  4. neutrons
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ _{3}Li^{7} + ^{1} _{1}P \Rightarrow  _{4}Be^{8} + _{0}^{0}\gamma (energy)$
As, the emitted particles is only energy and gamma photons radiation is nothing but energy so, option  C is correct.