T_A1/2 = ፒ_B
\(\frac{0.6931}{\lambda_{\mathrm{A}}}=\frac{1}{\lambda_{\mathrm{B}}}  \)
\(\lambda_{\mathrm{B}}=\frac{\lambda_{\mathrm{A}}}{0.6931}=1.44 \lambda_{\mathrm{A}}\)
Hence, B will decay at faster rate than A

B = ∆m x c²
∆m = \(\frac{B}{c^2}\)
N M_n + Z M_p - M(N,Z) = \(\frac{B}{c^2}\)
N (N,Z) = N M_n + ZM_p - \(\frac{B}{c^2}\)

r ∝ A^1/3
Surface Area = 4πr²
Hence, Surface Area ∝ A^2/3

Electric potential in n^th orbit
\(\mathrm{V} =\mathrm{V}_0 \ln \left(\frac{\mathrm{r}_{\mathrm{n}}}{\mathrm{r}_0}\right)  \)
\(=\mathrm{V}_0\left(\ln \mathrm{r}_{\mathrm{n}}-\ln \mathrm{r}_0\right)  \)
\(=\mathrm{V}_0 \ln \mathrm{r}_{\mathrm{n}}-\mathrm{V}_0 \ln \mathrm{r}_0\)
\(\left|\mathrm{F}_\epsilon\right|=\mathrm{e} \frac{\mathrm{dv}}{\mathrm{dr}}  =\mathrm{e} \frac{\mathrm{d}}{\mathrm{dr}}\left(\mathrm{V}_b / n \mathrm{r}_B-\mathrm{V}_0 / m \mathrm{r}_0\right)  \)
\(=\mathrm{c}\left(\frac{\mathrm{V}_0}{\mathrm{r}_n}-0\right)=\frac{\mathrm{eV}}{\mathrm{r}_{\mathrm{n}}} \)
Centripetal force = coulomb force
\(\frac{m v^2}{r_n}=\mathrm{c} \frac{V_0}{r_n} \Rightarrow v=\sqrt{\frac{e V_0}{m}}=\text { constant }\)
Angular momentum,
\(\mathrm{mvr}_n=\frac{\mathrm{nh}}{2 \pi}\)
\(\mathrm{m}, \mathrm{v}, \mathrm{h}, 2 \pi\) are constants
hence, r_n ∝ n