The Crrect Answer is

We know, $A=A_0{e}^{-\lambda t}$

For half life

$\frac{A_0}{2}=e^{-\lambda t_{1/2}}$

$\Rightarrow$$\lambda t_{1/2}$ = ln 2 .....(1)

And when radioactive element becomes ${\left(\frac{1}{8}\right)}^{th}$ of its initial value in 30 years

$\frac{A_0}{8}=A_0{e}^{-\lambda \times 30}\Rightarrow \lambda \times 30=\ln 8$

$\Rightarrow$ 30$\lambda =3\ln 2$

$\Rightarrow$$\lambda =\frac{3\ln 2}{30}$ .....(2)

Putting value of $\lambda$ in (1), we get

$\frac{3\ln 2}{30}\times t_{1/2}$ = ln 2

$\Rightarrow$$t_{1/2}$ = 10 years

The Correct Answer is Option (B)

Given, ($\tau$_1/2)_x = ($\tau$)_y

Here, $\tau$_1/2 = half-life period of radioactive element and $\tau$ = mean life period of radioactive element.

As we know the expression,

Half-life of the radioactive element x,

${\tau }_{1/2}=\frac{\ln (2)}{{\lambda }_x}$

Mean life of the radioactive element y,

$\tau =\frac{1}{{\lambda }_y}$

Substituting the values in Eq. (i), we get

$\frac{\ln 2}{{\lambda }_x}=\frac{1}{{\lambda }_y}\Rightarrow {\lambda }_x=0.693{\lambda }_y$

Initially they have same number of atoms,

$N_x=N_y=N_0$

As we know that,

Activity, $A=\lambda N$

As, ${\lambda }_x<{\lambda }_y\Rightarrow A_x<A_y$

Therefore, y will decay faster than x.

The Correct Answer is Option (D)

A$\to$ B $\to$C (stable)

Initially no. of atoms of B = 0 at t = 0, no. of atoms of B will starts increasing & reaches maximum value when rate of decay of B = rate of formation of B.

After that maximum value, no. of atoms will starts decreasing as growth & decay both are exponential functions, so best possible graph is (d).

The Correct Answer is Option (B)

$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{\frac{t}{t^{1/2}}}$

$\frac{N}{{10}^{10}}={\left(\frac{1}{2}\right)}^{\frac{30}{60}}$

$\Rightarrow N={10}^{10}\times {\left(\frac{1}{2}\right)}^{\frac{1}{2}}=\frac{{10}^{10}}{\sqrt{2}}\approx 7\times {10}^9$

The Correct Answer is Option (C)

A $\to$ B, B $\to$ C

$\frac{d{N}_B}{dt}=\lambda N_A-\lambda N_B$

$\frac{d{N}_B}{dt}=2\lambda N_{B_0}{e}^{-\lambda t}-\lambda N_B$

$e^{-\lambda t}\left(\frac{d{N}_B}{dt}+\lambda N_B\right)=2\lambda N_{B_0}{e}^{-\lambda t}\times e^{\lambda t}$

$\frac{d}{dt}(N_B{e}^{\lambda t})=2\lambda N_{B_0}$, on integrating

$N_B{e}^{\lambda t}=2\lambda t{N}_{B_0}+N_{B_0}$

$N_B=N_{B_0}[1+2\lambda t]e^{-\lambda t}$

$\frac{d{N}_B}{dt}=0$ at $-\lambda [1+2\lambda t)e^{-\lambda t}+2\lambda e^{-\lambda t}=0$

$N_{B_{max}}$$t=\frac{1}{2\lambda }$

The Correct Answer is Option (B)

(A) True, atom of each element emits characteristic spectrum.

(B) True, according to Bohr's postulates $mvr=\frac{nh}{2\pi }$ and hence electron resides into orbits of specific radius called stationary orbits.

(C) False, density of nucleus is constant.

(D) False, A free neutron is unstable decays into proton and electron and antineutrino.

(E) True, unstable nucleus show radioactivity.

The Correct Answer is Option (C)

N = N₀e^{$-$$\lambda$t}

N_d = N₀ $-$ N

N_d = N₀ (1 $-$ e^{$-$$\lambda$t})

$\frac{N_d}{N_0}=f=1-e^{-\lambda t}$$\Rightarrow$$\frac{df}{dt}=\lambda e^{-\lambda t}$

The Correct Answer is Option (D)

$\frac{3N_0}{4}=N_0{e}^{-\lambda t_1}$

$\frac{N_0}{2}=N_0{e}^{-\lambda t_2}$

$\ln (3/4)=-\lambda t_1$..... (i)

$\ln (1/2)=-\lambda t_2$ ..... (i)

$\ln (3/4)-\ln (1/2)=\lambda (t_2-t_1)$ ....(i)

$\mathrm{\Delta }t=\frac{\ln (3/2)}{\lambda }$

The Correct Answer is Option (C)

$k_{\alpha }+k_N=5.5$

$k=\frac{p^2}{2m}$

$\Rightarrow \frac{k_{\alpha }}{k_N}=\frac{180}{4}=45$

$\Rightarrow k_{\alpha }=\frac{45}{46}\times 5.5$ MeV

= 5.38 MeV

The Correct Answer is Option (D)

R = R₀e^{$-$$\lambda$t}

lnR = lnR₀$-$$\lambda$t

$-$$\lambda$is slope of straight line

$\lambda$ = $\frac{3}{20}$

$t_{1/2}=\frac{\ln 2}{\lambda }=4.62$

The Correct Answer is Option (A)

The given situation can be shown as


Here, radioactive material X is decayed into two particles Y and Z with their respective decay constant, $\lambda$_a and $\lambda$_b. It means that

$\because$$\lambda =\frac{\ln 2}{t_{1/2}}$

where, t_1/2(a) = 700 yr

and t_1/2(b) = 1400 yr

${\lambda }_a=\frac{\ln 2}{700}y{r}^{-1}$and${\lambda }_b=\frac{\ln 2}{1400}y{r}^{-1}$

$\because$${\lambda }_{total}={\lambda }_a+{\lambda }_b$

$=\left(\frac{\ln 2}{700}+\frac{\ln 2}{1400}\right)y{r}^{-1}=\ln 2\left(\frac{1}{700}+\frac{1}{1400}\right)y{r}^{-1}$

$=\left(\frac{3\ln 2}{1400}\right)y{r}^{-1}$

Suppose the initial number of radioactive nuclei was N₀.

$\therefore$$N=N_0{e}^{-\lambda t}$

where, N = number of nuclei present at time = t and N₀ = number of nuclei present at time = 0

$\Rightarrow \frac{N_0}{3}=N_0{e}^{-\lambda t}$ or $\frac{N_0}{3}=N_0{N}_0{e}^{-{\lambda }_{total}t}\Rightarrow \frac{1}{3}=e^{-{\lambda }_{total}t}$

Taking log on both the sides of above equation, we get

$\ln \left(\frac{1}{3}\right)=\ln (e^{-{\lambda }_{total}t})$

$\Rightarrow \ln \left(\frac{1}{3}\right)=-{\lambda }_{total}t$

$\Rightarrow 1.1=\frac{3\times 0.693}{1400}\times t$$\Rightarrow$t $\approx$ 740 yr

The Correct Answer is Option (D)

$T_{1/2}=20\Rightarrow \frac{\ln 2}{\lambda }=20$ min

$\Rightarrow \lambda =\frac{\ln 2}{20(min)}$

$\because$$N_t=N_0{e}^{-\lambda t}$

$\frac{N_t}{N_0}=e^{-\lambda t_1}\Rightarrow 0.67=e^{-\lambda t_1}$

$\Rightarrow \ln (0.67)=-\lambda t_1$

$\Rightarrow \ln \left(\frac{100}{67}\right)=\lambda t_1$

$\Rightarrow t_1=\frac{\ln \left(\frac{100}{67}\right)\times 20(min)}{(\ln 2)}$

Similarly, $t_2=\frac{\ln \left(\frac{100}{34}\right)\times 20(min)}{(\ln 2)}$

$t_2-t_1=19.57$ min $\approx 20$ min.

The Correct Answer is Option (B)

Activity, $A=\lambda N$

Where, $N=n{N}_A$

$N_A=6\times {10}^{23}$/mol

$=\frac{6\times {10}^{23}}{3.7\times {10}^{10}}$ Ci

Here, $A_0=\lambda N_0$

We know,

$T_{1/2}=\frac{\ln (2)}{\lambda }$

$\Rightarrow \lambda =\frac{\ln (2)}{T_{1/2}}$

$=\frac{\ln (2)}{2.7\times 3600\times 24}$

$\therefore$$A_0=\frac{\ln (2)}{2.7\times 3600\times 24}\times$$n\times N_A$

$=\frac{\ln (2)}{2.7\times 3600\times 24}\times \frac{1.5\times {10}^{-3}}{198}\times \frac{6\times {10}^{23}}{3.7\times {10}^{10}}$

$=366$ Ci

$\therefore$ Nearest answer is 357 Ci.

The Correct Answer is Option (D)

Let initial activity be A₀

A = A₀ e^{$-$$\lambda$t₁} ........(i)

$\frac{A}{5}$= A₀ e^{$-$$\lambda$t₂} .......(ii)

(i) $÷$ (ii)

5 = e^{$\lambda$(t₂ $-$ t₁)}

$\lambda$ = $\frac{\ln 5}{t_2-t_1}=\frac{1}{\tau }$

$\tau =\frac{t_2-t_1}{\ln 5}$

The Correct Answer is Option (D)

Let Half life of x = t

then half life of y = 2t

when 3 half life of y is completed then 6 half life of x is completed.

$\therefore$ Now x have =$\frac{N_1}{2^6}$ nuclei

and y have = $\frac{N_2}{2^3}$ nuclei

From question,

$\frac{N_1}{2^6}=\frac{N_2}{2^3}$

$\Rightarrow \frac{N_1}{N_2}=8$

The Correct Answer is Option (B)

T₁ = 10 sec

$\lambda$₁ = $\frac{\ln 2}{T_1}$

T₂ = 100 sec

$\lambda$₂ = $\frac{\ln 2}{T_2}$

$\lambda$_eq = $\frac{\ln 2}{T_{eq}}$

We know,

$\lambda$_eq = $\lambda$₁ + $\lambda$₂

$\Rightarrow$$\frac{\ln 2}{T_{eq}}$ =$\frac{\ln 2}{T_1}$ + $\frac{\ln 2}{T_2}$

$\Rightarrow$$\frac{1}{T_{eq}}$ = $\frac{1}{10}+\frac{1}{100}$

$\Rightarrow$T_eq = $\frac{100}{11}$ = 9 sec

The Correct Answer is Option (D)

$R=R_0{e}^{-\lambda t}$

$\ln R=ln{R}_0-\lambda t$

${\lambda }_A=\frac{6}{10}\Rightarrow T_A=\frac{10}{6}\ln 2$

${\lambda }_B=\frac{6}{5}\Rightarrow T_B=\frac{5\ln 2}{6}$

${\lambda }_C=\frac{2}{5}\Rightarrow T_C=\frac{5\ln 2}{2}$

$\frac{10}{6}:\frac{5}{6}:\frac{15}{6}::2:1:3$