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

AZX = 242He + 201e + Ai2iY
\(\frac{N_i}{Z_i}=\frac{(A_i-Z_i)}{Z_i} =\frac{A-8-(Z-6)}{Z-6}=\frac{A-Z-2}{Z-6}\)

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 - \(\frac{B}{c^2}\)

\(\frac{BE}{A}=\frac{\Delta\times931MeV}{V}=\frac{0.042 \times931}{7}\)
= 5.586 = 5.6 MeV

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

\(r \propto A^{\frac{1}{3}}  \)
\(\frac{r_{\mathrm{Cu}}}{\mathrm{r}_{\mathrm{Al}}}=\frac{\mathrm{A}_{\mathrm{Cu}}^\frac{1}{3}}{\mathrm{~A}_{\mathrm{Al}}^{\frac{1}{3}}}=\frac{4}{3}  \)
\(\mathrm{r}_{\mathrm{Cu}}=\frac{4}{3} \times 3.6 \mathrm{~F}=4.8 \mathrm{~F}\)

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

\(\frac{1}{\lambda}=\mathrm{RZ}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]  \)
\(\frac{1}{\lambda}=\mathrm{RZ} ^2\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3}{4} \mathrm{RZ}^2  \)
\(\lambda \propto \frac{1}{Z^2}  \)
\(\lambda_{\mathrm{Li}}: \lambda_{\mathrm{H} c}: \lambda_{\mathrm{H}}=\frac{1}{9}: \frac{1}{4}: \frac{1}{1}=4: 9: 36\)

Cathode rays are stream of negatively charged electron.

r_n ∞ n²
r₁: r₂: r₃ = 1: 4: 9

\(V_{ionisation}=\frac{13.6}{n^2}Z^2 volt\)
\(Z=\sqrt\frac{V\times n^2}{13.6}=\sqrt\frac{122.4 \times I^2}{13.6}=\sqrt{9}=3\)

\(L=\frac{nh}{2\pi}=\frac{4h}{2\pi}=\frac{2h}{\pi}\)