Class 12th

$\frac{\mathrm{N}}{{\mathrm{N}}_0}=\frac{1}{16}={\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{{\mathrm{T}}_{1/2}}}$

$\mathrm{t}=4{\mathrm{T}}_{1/2}=20*4=80\text{minutes}$

$?\frac{1}{\lambda }=\mathrm{R}\left(\frac{1}{{\mathrm{n}}_2^2}-\frac{1}{{\mathrm{n}}_1^2}\right)$

$\Rightarrow \frac{1}{\lambda }=\mathrm{R}\left(\frac{1}{2^2}\right)$

$\Rightarrow \lambda =\frac{4}{\mathrm{R}}$

$\frac{1}{{\lambda }^{\mathrm{\text{'}}}}=\mathrm{R}\left(\frac{1}{4^2}\right)$

$\Rightarrow {\lambda }^{\mathrm{\text{'}}}=\frac{16}{\mathrm{R}}$

$\Rightarrow {\lambda }^{\mathrm{\text{'}}}=4\lambda$

Speed of light in air is

$\mathrm{C}=\frac{\mathrm{x}}{{\mathrm{t}}_1}$

Speed of light in another denser medium.

${\mathrm{C}}_2=\frac{10\mathrm{x}}{{\mathrm{t}}_2}$

$\Rightarrow \mu =\frac{C}{C_2}=\frac{x}{t_1}*\frac{t_2}{10x}$

$\Rightarrow \mu =\frac{{\mathrm{t}}_2}{10{\mathrm{t}}_1}$

(d) $\mathrm{\Delta }\theta =\frac{\lambda }{\mathrm{d}}$

$\mathrm{\Delta }\theta$ is proportional to $\lambda$ but independent of $D$ .

(c) $\mathrm{e}\mathrm{V}=$ Energy of electron

for minimum wavelength, maximum loss of energy

$\mathrm{e}\mathrm{V}=\left(\frac{\mathrm{h}\mathrm{c}}{\lambda }\right)$

$\lambda \propto \left(\frac{1}{\mathrm{V}}\right)$

31. (b) Incident energy $=2.20\mathrm{e}\mathrm{V}$

If $?<2.20\mathrm{e}\mathrm{V}$  electron will emit.

$?>2.20\mathrm{e}\mathrm{V}\mathrm{}$ No electron emission

Only caesium will emit electron

$\mathrm{V}=\frac{1}{4\pi {?}_0}\cdot \frac{\mathrm{Q}}{\mathrm{R}}$

$\therefore \mathrm{V}\propto \frac{1}{\mathrm{R}}$

$\therefore$  Potential is more on smaller sphere.

$\mathrm{B}=\frac{\mathrm{E}}{\mathrm{C}}=\frac{48}{3*{10}^8}=16*{10}^{-8}$

$=1.6*{10}^{-7}\mathrm{T}$

$\mathrm{y}=\left(\mathrm{n}\lambda \right)\left(\frac{\mathrm{D}}{\mathrm{d}}\right)$

${\mathrm{n}}_1{\lambda }_1={\mathrm{n}}_2{\lambda }_2$

(8) $\left(600\mathrm{n}\mathrm{m}\right)={\mathrm{n}}_2\left(400\mathrm{n}\mathrm{m}\right)$

${\mathrm{n}}_2=12$

(a) Radio wave (ii) $\approx 102\mathrm{m}$   (ii)
(b) Microwave $\approx$ (iii) $10.2\mathrm{m}$   (iii)

(c) Infrared radiations $\approx$ (iv) $10.4\mathrm{m}$   (iv)

(d) $\mathrm{X}$ - ray (i) $\approx \text{?}=10.10\mathrm{m}$   (i)

(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)

In half wave rectification $\Rightarrow {\mathrm{f}}_{\text{output}}=60\mathrm{H}\mathrm{z}$

$\mathrm{E}=\mathrm{P}*\mathrm{t}=100*{10}^3*3600$

$=36*{10}^7\mathrm{J}$

$\left. [{\mathrm{M}\mathrm{L}\mathrm{T}}^{-2}{\mathrm{A}}^{-2}\right]=$ Magnetic permeability