$$\eqalign{ & {\left( {{T_{\frac{1}{2}}}} \right)_A} = {\left( {{t_{{\text{mean}}}}} \right)_B} \cr & \Rightarrow \frac{{0.693}}{{{\lambda _A}}} = \frac{1}{{{\lambda _B}}} \cr & \Rightarrow {\lambda _A} = 0.693{\lambda _B} \cr & {\text{or}}\,\,{\lambda _A} < {\lambda _B} \cr} $$
Also rate of decay $$ = \lambda N$$
Initially number of atoms $$\left( N \right)$$  of both are equal but since $${\lambda _B} > {\lambda _A},$$   therefore $$B$$ will decay at a faster rate than $$A$$

\[_n^mX{\xrightarrow{\alpha }_{n - 2}}{X^{m - 4}}{\xrightarrow{{2\beta }}_n}{X^{m - 4}}\]

$$\eqalign{ & {A_1} = \left( {\lambda {N_0}} \right){e^{ - \lambda {t_1}}}\,......\left( {\text{i}} \right) \cr & {A_2} = \left( {\lambda 2{N_0}} \right){e^{ - \lambda {t_2}}}\,......\left( {{\text{ii}}} \right) \cr & {t_1} - {t_2} = \frac{1}{\lambda }\ell n\left( {\frac{{{A_2}}}{{2{A_1}}}} \right) = \ell n\left( {\frac{{{A_2}}}{{2{A_1}}}} \right) \cr} $$

$$\eqalign{ & \frac{{d{N_A}}}{{dt}} = - {\lambda _1}{N_A},\frac{{d{N_B}}}{{dt}} = 2{\lambda _1}{N_A} - {\lambda _2}{N_B}, \cr & {N_B} = {\text{maximum}} \Rightarrow \frac{{d{N_B}}}{{dt}} = 0 \cr & \Rightarrow 2{\lambda _1}{N_A} = {\lambda _2}{N_B}_{_{\max }} \cr & \Rightarrow {N_B}_{_{\max }} = \frac{{2{\lambda _1}}}{{{\lambda _2}}}{N_A} \cr & \Rightarrow {N_B}_{_{\max }} = \frac{{2{\lambda _1}}}{{{\lambda _2}}}{N_0}{e^{ - {\lambda _1}t}} = 2 \cr} $$

$$_{92}{U^{238}}{ \to _{92}}T{h^{238}}{ + _2}H{e^4}$$
According to law of conservation of linear momentum, we have.
$$\left| {{P_{Th}}} \right| = \left| {{P_{He}}} \right| = P$$
⇒ As, kinetic energy of an element,
$$KE = \frac{{{P^2}}}{{2m}}$$
where, $$m$$ is mass of an element.
Thus, $$KE \propto \frac{1}{M}$$
So, $${M_{He}} < {M_{Th}} \Rightarrow {K_{He}} > {K_{Th}}$$

$$\eqalign{ & {N_1} = {N_0}{e^{ - 10\lambda t}}\,{\text{and}}\,{N_2} = {N_0}{e^{ - \lambda t}}. \cr & \therefore \frac{{{N_1}}}{{{N_2}}} = \frac{{{e^{ - 10\lambda t}}}}{{{e^{ - \lambda t}}}} = \frac{1}{{{e^{9\lambda t}}}} \cr & {\text{Given}}\,\,\frac{{{N_1}}}{{{N_2}}} = \frac{1}{e}; \cr & \therefore \frac{1}{{{e^{9\lambda t}}}} = \frac{1}{e} \cr & {\text{or,}}\,\,9\lambda t = 1\,\,{\text{or}}\,\,t = \left( {\frac{1}{{9\lambda }}} \right) \cr} $$

Isotopes : $$A\,\& \,C$$  have same number of protons.

$${k_\alpha } = \frac{M}{{M + 4}}Q = \frac{{222}}{{226}} \times 4.87 = 4.78\,MeV$$

Let number of atoms in $$X = {N_x}$$
Number of atoms in $$Y = {N_y}$$
By question
$$\eqalign{ & \frac{{{N_x}}}{{{N_y}}} = \frac{1}{{15}} \cr & \therefore {\text{Part}}\,{\text{of}}\,{N_x} = \frac{1}{{16}}\left( {{N_x} + {N_y}} \right) = \frac{1}{{{2^4}}}\left( {{N_x} + {N_y}} \right) \cr} $$
So, total 4 half lives are passed, so, age of rock is $$4 \times 50 = 200\,{\text{years}}$$

The given nuclear reaction can be written as
$$_Z{X^A}{ + _0}{n^1}{ \to _3}L{i^7}{ + _2}H{e^4}$$
Conservation of mass number gives,
$$A + 1 = 7 + 4 \Rightarrow A = 10$$
Conservation of charge number/Atomic No. gives,
$$Z + 0 = 2 + 3 \Rightarrow Z = 5$$
Hence, $$Z = 5,A = 10$$    corresponds to boron $$\left( {_5{B^{10}}} \right).$$