Nuclei Set-4

Q1. A radioactive sample decays so that its activity becomes one-eighth of the initial activity after $30\ \text{days}$. Using $N(t)=N_0 2^{-t/T_{1/2}}$, the half-life $T_{1/2}$ is:

Q2. A nucleus with atomic mass $M_P=226.0254\ \mathrm{u}$ undergoes $\alpha$-decay to a daughter of mass $M_D=222.0176\ \mathrm{u}$. Mass of the $\alpha$-particle is $M_\alpha=4.0015\ \mathrm{u}$. Using $Q=(M_P-(M_D+M_\alpha))c^2$ and $1\ \mathrm{u}=931.5\ \mathrm{MeV}/c^2$, the $\alpha$-decay is:

Q3. Parent nucleus $P$ ($T_{1/2,P}=10\ \text{days}$) decays to daughter $D$ ($T_{1/2,D}=1\ \text{day}$). Initially $N_P(0)=N_0$ and $N_D(0)=0$. If $\lambda=\ln2/T_{1/2}$, the time $t_{\max}$ at which the number of daughter nuclei is maximum is given by $t_{\max}=\dfrac{1}{\lambda_D-\lambda_P}\ln\dfrac{\lambda_D}{\lambda_P}$. Numerically $t_{\max}\approx$:

Q4. Assertion (A): The fusion of two deuterons to form one $^4$He nucleus releases more energy per nucleon than a typical fission of $^{235}$U into two roughly equal fragments.
Reason (R): To achieve fusion the interacting nuclei must overcome (or quantum-tunnel through) the Coulomb barrier between their positive charges.

Q5. A proton-rich nuclide $X$ has atomic mass $M_X=60.00100\ \mathrm{u}$ and its daughter $Y$ has atomic mass $M_Y=60.00010\ \mathrm{u}$. Using $1\ \mathrm{u}=931.5\ \mathrm{MeV}/c^2$ and $2m_ec^2=1.022\ \mathrm{MeV}$, which decay mode(s) are energetically allowed for $X$?

Q6. The masses are: proton $m_p=1.007276\ \text{u}$, neutron $m_n=1.008665\ \text{u}$ and the $^4_2\text{He}$ nucleus has mass $m\left(^4\text{He}\right)=4.001506\ \text{u}$. Using $1\ \text{u}=931.5\ \text{MeV}/c^2$, the binding energy per nucleon of $^4_2\text{He}$ is approximately:
(A) $6.50\ \text{MeV}$
(B) $7.28\ \text{MeV}$
(C) $7.07\ \text{MeV}$
(D) $7.50\ \text{MeV}$
Answer: C
Explanation:

1. For $^4_2\text{He}$, $Z=2$ and $N=2$.
2. Mass defect: $$\Delta m = 2m_p+2m_n - m(^4\text{He}).$$
3. Compute: $$2m_p+2m_n = 2(1.007276)+2(1.008665)=4.031882\ \text{u}.$$
4. Then $$\Delta m = 4.031882-4.001506=0.030376\ \text{u}.$$
5. Total binding energy: $$E_b = \Delta m \, c^2 = 0.030376\times 931.5 \approx 28.29\ \text{MeV}.$$
6. Binding energy per nucleon (4 nucleons): $$\frac{E_b}{4}\approx \frac{28.29}{4}\approx 7.07\ \text{MeV}.$$

Therefore Option C is correct.

Q7. A nucleus of mass number $A=220$ emits an $\alpha$-particle with total available energy (Q-value) $Q=5.0\ \text{MeV}$. Assuming nuclear masses are proportional to mass numbers and using conservation of momentum, the kinetic energy carried by the emitted $\alpha$-particle is approximately:
(A) $4.91\ \text{MeV}$
(B) $5.00\ \text{MeV}$
(C) $4.75\ \text{MeV}$
(D) $4.59\ \text{MeV}$
Answer: A
Explanation:

1. After $\alpha$ emission: daughter mass number $A_D=220-4=216$.
2. Let masses be proportional to mass numbers: $m_\alpha \propto 4$, $m_D \propto 216$.
3. By conservation of momentum, momenta are equal: $p_\alpha=p_D$.
4. Kinetic energies: $$K_\alpha=\frac{p^2}{2m_\alpha},\quad K_D=\frac{p^2}{2m_D}.$$
5. Ratio: $$\frac{K_D}{K_\alpha}=\frac{m_\alpha}{m_D}=\frac{4}{216}.$$
6. Total Q-value: $$Q=K_\alpha+K_D=K_\alpha\left(1+\frac{4}{216}\right) =K_\alpha\left(\frac{220}{216}\right).$$
7. Hence $$K_\alpha=Q\frac{216}{220}=5.0\times\frac{216}{220}\approx 4.91\ \text{MeV}.$$

So Option A is correct.

Q8. In a decay chain $X \xrightarrow{\lambda_1} Y \xrightarrow{\lambda_2} \text{stable}$, the decay constants are $\lambda_1=0.02\ \text{s}^{-1}$ and $\lambda_2=0.10\ \text{s}^{-1}$. Initially $N_X(0)=N_0$ and $N_Y(0)=0$. At what time $t$ is the number of $Y$ nuclei maximum?
(A) $12.5\ \text{s}$
(B) $18.0\ \text{s}$
(C) $25.0\ \text{s}$
(D) $20.12\ \text{s}$
Answer: D
Explanation:

1. For $N_Y(0)=0$, the number of $Y$ nuclei is: $$N_Y(t)=\frac{N_0\lambda_1}{\lambda_2-\lambda_1}\left(e^{-\lambda_1 t}-e^{-\lambda_2 t}\right).$$
2. Maximum occurs when $\dfrac{dN_Y}{dt}=0$.
3. Differentiate and set to zero: $$-\lambda_1 e^{-\lambda_1 t}+\lambda_2 e^{-\lambda_2 t}=0 \Rightarrow \lambda_1 e^{-\lambda_1 t}=\lambda_2 e^{-\lambda_2 t}.$$
4. Rearranging: $$e^{(\lambda_2-\lambda_1)t}=\frac{\lambda_2}{\lambda_1}.$$
5. So $$t=\frac{1}{\lambda_2-\lambda_1}\ln\left(\frac{\lambda_2}{\lambda_1}\right).$$
6. Substitute values: $$t=\frac{1}{0.10-0.02}\ln\left(\frac{0.10}{0.02}\right) =\frac{1}{0.08}\ln(5).$$
7. Numerically, $\ln(5)\approx 1.609$: $$t\approx \frac{1.609}{0.08}=20.12\ \text{s}.$$

Thus Option D is correct.

Q9. A projectile of mass number $1$ (mass $m$) strikes a stationary target of mass number $14$ (mass $M$) and induces an endothermic nuclear reaction with $Q=-4.0\ \text{MeV}$. Neglect internal excitations and take nucleon masses proportional to mass numbers. The minimum kinetic energy (threshold) required in the laboratory frame is closest to:
(A) $4.00\ \text{MeV}$
(B) $4.29\ \text{MeV}$
(C) $3.73\ \text{MeV}$
(D) $4.58\ \text{MeV}$
Answer: B
Explanation:

1. For an endothermic reaction with negative Q-value, the lab-frame threshold energy is (neglecting internal excitations): $$K_{\text{th}} = -Q\left(1+\frac{m}{M}\right).$$
2. Here $Q=-4.0\ \text{MeV}$, so $-Q=4.0\ \text{MeV}$.
3. Using proportionality of mass to mass number: $$\frac{m}{M}=\frac{1}{14}.$$
4. Substitute: $$K_{\text{th}}=4.0\left(1+\frac{1}{14}\right) =4.0\left(\frac{15}{14}\right)=\frac{60}{14}\approx 4.29\ \text{MeV}.$$

So the closest value is Option B.

Q10. Assertion (A): A nucleus can undergo positron emission ($\beta^+$ decay) only if the mass of the parent neutral atom exceeds the mass of the daughter neutral atom by at least $2m_e c^2$.
Reason (R): Emission of a positron requires creation of a particle with rest energy $m_e c^2$, and because the daughter neutral atom has one fewer electron than the parent, the atomic-mass comparison leads to a minimum energy cost of $2m_e c^2$.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Answer: A
Explanation:

1. For $\beta^+$ decay, the Q-value using neutral atomic masses is: $$Q=[M_{\text{parent}}-M_{\text{daughter}}-2m_e]c^2.$$
2. For decay to be energetically allowed, $Q\ge 0$: $$M_{\text{parent}}-M_{\text{daughter}}\ge 2m_e.$$ So Assertion (A) is true.
3. Reason (R) explains the origin of the $2m_e c^2$ term:
   • One $m_ec^2$ for creating the positron,
   • Another $m_ec^2$ because comparing neutral atoms effectively accounts for removal of one electron in the daughter.
4. Hence (R) correctly explains why the mass difference must be at least $2m_e c^2$.
   So Option A is correct.

Q11. A nucleus of $^{20}_{10}\mathrm{Ne}$ has measured nuclear mass $m_{\mathrm{nucl}}=19.96348\ \mathrm{u}$. Given $m_p=1.007276\ \mathrm{u}$, $m_n=1.008665\ \mathrm{u}$ and $1\ \mathrm{u}=931.5\ \mathrm{MeV}/c^2$, calculate the binding energy per nucleon. (Use $\Delta m = Zm_p+Nm_n-m_{\mathrm{nucl}}$ and $E_b=\Delta m\,c^2$.)

Q12. Two radioactive isotopes X and Y are present in equal numbers initially. Their half-lives are $t_{1/2,X}=2\ \mathrm{h}$ and $t_{1/2,Y}=8\ \mathrm{h}$. After what time $t$ (in hours) will their activities become equal? (Use $\lambda=\ln2/t_{1/2}$.)

Q13. For a nuclear reaction in the lab frame with target at rest the threshold kinetic energy of the projectile is given by $E_{\mathrm{th}}=-Q\left(1+\dfrac{m}{M}\right)$ for negative $Q$, where $m$ is projectile mass and $M$ the target mass. If a proton ($m=1\ \mathrm{u}$) strikes a $^{12}\mathrm{C}$ target ($M=12\ \mathrm{u}$) and the reaction has $Q=-2.00\ \mathrm{MeV}$, the minimum incident proton kinetic energy required is approximately:

Q14. A parent nucleus P (half-life $T_P=30\ \text{days}$) decays to daughter D (half-life $T_D=6\ \text{h}$). Initially only the parent is present. For the decay chain P→D→stable the activity ratio evolves as

where $\lambda=\ln2/T_{1/2}$. Approximately how long (in hours) will it take for $A_D$ to reach $90\%$ of $A_P$?

Q15. Assertion (A): For $\beta^-$ decay the $Q$-value expressed in terms of atomic masses is $Q=[M(Z,A)-M(Z+1,A)]c^2$, while for $\beta^+$ decay it is $Q=[M(Z,A)-M(Z-1,A)-2m_e]c^2$. Reason (R): Because atomic masses include the electrons bound to nuclei, a $\beta^-$ emitted electron is already accounted for in the daughter atomic mass, whereas $\beta^+$ requires creation of a positron and removal of one atomic electron, costing an extra $2m_ec^2$. Which is correct?