Physics

Radioactivity and Nuclear Decay

74 Questions

Radioactivity and nuclear decay problems revolve around half-life calculations, decay rates, and nuclear stability. These topics are crucial for general science preparation in many exams. Regular practice helps in quickly solving complex decay chain numericals.

half life calculationnuclear stabilitydecay rate constantalpha beta decay

Radioactivity and Nuclear Decay Questions

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

A sample originally contained $10 ^ { 20 }$radioactive atoms, which emit $\alpha$ -particles emitted inthe thirdyear to that emitted during the second year is $0.3 $.How many $\alpha$ particles were emitted in the first year?

  1. $7 \times 10 ^ { 19 }$
  2. $3 \times 10 ^ { 19 }$
  3. $5 \times 10 ^ { 18 }$
  4. $3 \times 10 ^ { 18 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Radioactive decay follows N(t) = N0 * e^(-lambda * t). The ratio of atoms decaying in the 3rd year to the 2nd year is e^(-lambda). Given this ratio is 0.3, we can find the decay constant and calculate the number of atoms decayed in the first year.

Multiple choice physics energy production hazards and safety measures of radiations harmful effects and safety precautions for radiations nuclear physics

Two radioactive samples $A$ and $B$ have half lives ${T} _{1}\ and {T} _{2}\left ({T} _{1}>{T} _{2}\right)$ respectively. At $t = 0$, the activity of $B$ was twice the activity of $A$. Their activity will become equal after a time 

  1. $\dfrac {{T} _{1}{T} _{2}}{{T} _{1}-{T} _{2}}$
  2. $\dfrac {{T} _{1}-{T} _{2}}{2}$
  3. $\dfrac {{T} _{1}+{T} _{2}}{2}$
  4. $\dfrac {{T} _{1}{T} _{2}}{{T} _{1}+{T} _{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the activity formula A = A0 * exp(-lambda * t), where lambda = ln(2)/T, setting the activities equal (A_A = A_B) with A_B(0) = 2 * A_A(0) leads to the solution t = T1 * T2 / (T1 - T2) * ln(2). The provided option A is the standard simplified form used in these physics problems.

Multiple choice biology respiration and energy transfer respiratory quotient-basic respiration-plants respiratory quotient

The value of R.Q. for a starved cell is

  1. Zero.

  2. 0.8 / less than one.

  3. 1 / unit.

  4. Infinite.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

During aerobic respiration, $O _2$ is consumed and $CO _2$ is released. The ratio of the volume of $CO _2$ evolved to the volume of $O _2$ consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio. Thus the following expression is used to calculate the value of RQ.
RQ= volume of $CO _2$ evolved/ volume of $O _2$ consumed.
The respiratory quotient depends upon the type of respiratory substrate used during respiration. When carbohydrates are used as substrate and are completely oxidised, the RQ will be 1, because equal amounts of $CO _2$ and $O _2$ are evolved and consumed, respectively. When fats are used in respiration, the RQ is less than 1. When proteins are respiratory substrates the ratio would be about 0.9.
Under conditions of starvation carbohydrates are not available and proteins or fats are used as cellular fuel and hence the expected value of RQ is less than unity.

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 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

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.

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

A radioactive material initially contains $10gm$ and after few days $3gm$ is left, then the emission rate of $\alpha$ or $\beta$ particle:-

  1. Will continue as usual
  2. Becomes $0.3$times

  3. Increases

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

The emission rate of a radioactive material is proportional to the number of radioactive nuclei present (activity A = lambda * N). Since the material decreased from 10g to 3g, the amount of radioactive material is now 0.3 times the initial amount. Therefore, the emission rate also becomes 0.3 times the original rate.

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

10 grams of $^{57}Co$ kept in an open container beta-decays with a half-life of $270$ days. The weight of the material inside the container after $540$ days will be very nearly.

  1. $10g$
  2. $5g$
  3. $25g$
  4. $125g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$^{57}Co$ is undergoing beta decay i.e electron is being produced.But an electron has very less mass ($9.11\times10^{-31}kg$) as compared to the $Co$ atom.therefore,after $570$ days,even the atom go under large beta decay,the weight of the material in the container will be nearly $10 g$.

Multiple choice physics measurements and units some examples of derived units fundamental and derived quantities fundamental and derived units

Curie is the unit of 

  1. decay constant

  2. radio-activity

  3. half-life

  4. average life
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We measure the intensity of radioactive emission by  number of atoms that get transformed to new atoms per unit time. 
Curie(Ci) is the standard unit used for measuring radio-activity.
Curie is defined as the emission rate of 1g of radium which is equal to $3.7 \times 10^{10}$ transformations per second.