Mix Examples-Nuclei Questions in English

(C) The number of undecayed radioactive atoms at time $t$ is given by $N = N_0 e^{-\lambda t}$.
The number of decayed atoms $N'$ at time $t$ is the difference between the initial number of atoms $N_0$ and the remaining atoms $N$:
$N' = N_0 - N = N_0 - N_0 e^{-\lambda t} = N_0(1 - e^{-\lambda t})$.
At $t = 0$,$N' = N_0(1 - e^0) = 0$.
As $t \to \infty$,$N' \to N_0$.
This represents a curve that starts from the origin $(0,0)$ and increases exponentially,approaching the horizontal asymptote $N' = N_0$. This matches the graph shown in option $C$.

(A) According to the law of conservation of linear momentum,the initial momentum of the system is zero because the parent nucleus is at rest.
Let $M$ be the mass of the parent nucleus,$m_{\alpha} = 4$ be the mass of the $\alpha$-particle,and $m_d = A - 4$ be the mass of the daughter nucleus.
Since the initial momentum is zero,the final momentum must also be zero:
$P_{initial} = P_{final}$
$0 = m_d \upsilon' + m_{\alpha} \upsilon$
Here,$\upsilon'$ is the velocity of the daughter nucleus and $\upsilon$ is the velocity of the $\alpha$-particle.
$(A - 4) \upsilon' + 4 \upsilon = 0$
$(A - 4) \upsilon' = -4 \upsilon$
$\upsilon' = -\frac{4 \upsilon}{A - 4}$
The magnitude of the velocity of the daughter nucleus is $\frac{4 \upsilon}{A - 4}$.

(C) According to the law of conservation of linear momentum,the total initial momentum of the system is zero because the nucleus is initially at rest.
Let $M$ be the mass of the parent nucleus,$m_{\alpha}$ be the mass of the $\alpha$-particle,and $m_d$ be the mass of the daughter nucleus.
Since mass is proportional to the mass number,we have $M \propto A$,$m_{\alpha} \propto 4$,and $m_d \propto (A - 4)$.
Initial momentum $P_i = 0$.
Final momentum $P_f = m_d v' - m_{\alpha} v = 0$,where $v'$ is the velocity of the daughter nucleus.
Substituting the mass numbers: $(A - 4)v' - 4v = 0$.
Therefore,$(A - 4)v' = 4v$.
Solving for $v'$,we get $v' = \frac{4v}{A - 4}$.

(A) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
From this,we have $\lambda = \frac{hc}{E}$,which implies $\frac{\lambda_N}{\lambda_A} = \frac{E_A}{E_N}$.
Here,$E_N$ is the energy of the photon emitted during nuclear de-excitation,which is typically in the order of $MeV$ $(10^6 \ eV)$.
$E_A$ is the energy of the photon emitted during atomic de-excitation,which is typically in the order of a few $eV$ (e.g.,$1-10 \ eV$).
Therefore,the ratio $\frac{\lambda_N}{\lambda_A} = \frac{E_A}{E_N} \approx \frac{1 \ eV}{10^6 \ eV} = 10^{-6}$.

$(A)$ True: Atoms of each element emit a characteristic spectrum due to electronic transitions.
$(B)$ True: According to Bohr's postulate, electrons revolve in specific orbits where angular momentum is quantized $(mvr = \frac{nh}{2\pi})$, known as stationary orbits.
$(C)$ False: The density of nuclear matter is independent of the mass number $(A)$ and is approximately constant for all nuclei $(\approx 2.3 \times 10^{17} \, kg/m^3)$.
$(D)$ False: A free neutron is unstable and decays into a proton, an electron, and an antineutrino $(n \rightarrow p + e^- + \bar{\nu}_e)$, whereas a free proton is stable.
$(E)$ True: Radioactivity is a spontaneous process resulting from the instability of atomic nuclei.
Therefore, statements $A, B,$ and $E$ are correct.

(C) The correct matches are as follows:
$1$. The binding energy per nucleon for ${ }^{56} Fe$ is approximately $8.75 \,MeV$, which is the most stable nucleus. Thus, $a \rightarrow (iii)$.
$2$. The energy of $\alpha$-particles used in the Geiger-Marsden experiment is typically around $5.5 \,MeV$. Thus, $b \rightarrow (i)$.
$3$. The energy of a photon of visible light ranges from approximately $1.6 \,eV$ to $3.2 \,eV$, so $2 \,eV$ is a representative value. Thus, $c \rightarrow (iv)$.
$4$. The energy released in the fission of a single uranium nucleus is approximately $200 \,MeV$. Thus, $d \rightarrow (ii)$.
Therefore, the correct sequence is $a(iii), b(i), c(iv), d(ii)$.

(C) Statement $(A)$ is correct according to Bohr's postulate: $L = n\hbar$.
Statement $(B)$ is correct because nuclear forces are short-range and do not follow the inverse square law like gravitational or electrostatic forces.
Statement $(C)$ is correct as nuclear forces depend on the relative spin orientations of nucleons.
Statement $(D)$ is incorrect because nuclear forces are non-central forces.
Statement $(E)$ is correct because a lower packing fraction indicates a higher binding energy per nucleon,leading to greater nuclear stability.
Therefore,statements $(A), (B), (C),$ and $(E)$ are correct.

$(A)$ The energy of the system increases in $(p)$ (charging a capacitor), $(q)$ (adiabatic compression increases internal energy), and $(t)$ (induced current causes heating).
$(B)$ In $(q)$, mechanical work done on the piston increases the random kinetic energy of gas molecules.
$(C)$ In $(s)$, the mass-energy equivalence $E = mc^2$ implies that the decrease in mass is converted into the kinetic energy of the fragments (mechanical energy).
$(D)$ In $(s)$, nuclear fission results in a mass defect, meaning the mass of the system decreases.
Therefore, the correct matching is: $A-(p, q, t), B-q, C-s, D-s$.

(C) In $\alpha$ decay,the mass number decreases by $4$ and the atomic number decreases by $2$. This matches process $2$ $({ }_{92}^{238} U \rightarrow{ }_{90}^{234} Th + { }_{2}^{4} He)$. So,$P-2$.
In $\beta^{+}$ decay,the mass number remains unchanged while the atomic number decreases by $1$. This matches process $1$ $({ }_{8}^{15} O \rightarrow{ }_{7}^{15} N + { }_{+1}^{0} e)$. So,$Q-1$.
In nuclear fission,a heavy parent nucleus splits into two smaller,roughly equal fragments. This matches process $4$ $({ }_{94}^{239} Pu \rightarrow{ }_{57}^{140} La + \dots)$. So,$R-4$.
In proton emission,a proton is ejected,so both the mass number and atomic number decrease by $1$. This matches process $3$ $({ }_{83}^{185} Bi \rightarrow{ }_{82}^{184} Pb + { }_{1}^{1} H)$. So,$S-3$.
Therefore,the correct matching is $P-2, Q-1, R-4, S-3$.

(A) Let the mass number of the parent nucleus be $A = 208$. The mass of an alpha particle is $m_{\alpha} \approx 4$ atomic mass units.
Two alpha particles are emitted,each with kinetic energy $E$. The total kinetic energy of the alpha particles is $K_{\alpha} = 2E$.
The momentum of each alpha particle is $p_{\alpha} = \sqrt{2 m_{\alpha} E}$.
Since the parent nucleus is at rest,the total momentum must be zero: $\vec{p}_{d} + \vec{p}_{\alpha 1} + \vec{p}_{\alpha 2} = 0$,where $\vec{p}_{d}$ is the momentum of the daughter nucleus.
Assuming the alpha particles are emitted in opposite directions,the net momentum of the alpha particles is zero,so the daughter nucleus remains at rest $(K_{d} = 0)$.
However,if they are emitted at an angle,the daughter nucleus must recoil. But the question asks for the total kinetic energy released in the disintegration,which is the sum of the kinetic energies of all products.
The total energy released $Q$ is the sum of kinetic energies: $Q = K_{\alpha 1} + K_{\alpha 2} + K_{d}$.
Given the symmetry and the conservation of momentum,the total kinetic energy is the sum of the energies of the particles: $K_{total} = E + E + K_{d}$.
For a nucleus of mass $208$ emitting two alpha particles ($8$ units),the daughter nucleus has mass $200$. The recoil energy $K_{d} = \frac{p_{d}^2}{2M_{d}}$.
Using conservation of momentum for the system,the total kinetic energy is $K_{total} = 2E(1 + \frac{m_{\alpha}}{M_{d}}) = 2E(1 + \frac{4}{200}) = 2E(1 + \frac{1}{50}) = 2E(\frac{51}{50}) = \frac{51E}{25}$.