Physics Ncert Solutions Class 12th

The reaction is X²? → Y¹²? + Z¹²?
Binding energies per nucleon are: X=7.6 MeV, Y=8.5 MeV, Z=8.5 MeV.
Gain in binding energy (Q) = (Binding energy of products) - (Binding energy of reactants)
Q = (120 * 8.5 + 120 * 8.5) - (240 * 7.6) MeV
Q = (2 * 120 * 8.5) - (240 * 7.6) MeV = 2040 - 1824 = 216 MeV.

In n-type semiconductor majority charge carriers are e- and P type semiconductor majority charge carriers are holes.

$\mathrm{I}={\mathrm{n}\mathrm{e}\mathrm{A}\mathrm{V}}_{\mathrm{d}}=$ neA $\left(\mu \mathrm{E}\right)$

${\mu }_{\mathrm{e}}>{\mu }_{\mathrm{h}}\Rightarrow {\mathrm{I}}_{\mathrm{e}}>{\mathrm{I}}_{\mathrm{h}}$

${\mathrm{Y}}_{\text{copper}}<{\mathrm{Y}}_{\text{steel}}$

$\mathrm{E}=\frac{\mathrm{K}\mathrm{p}}{{\mathrm{r}}^3}\sqrt[]{1+3{\mathrm{c}\mathrm{o}\mathrm{s}}^2?\theta }$

$\therefore \mathrm{E}\propto \frac{1}{{\mathrm{r}}^3}$

First excited state $\Rightarrow n=2$ $$

${\mathrm{T}}_1=13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}=-\frac{13.6}{4}\mathrm{e}\mathrm{V}$

Second excited state $\Rightarrow \mathrm{n}=3$ $$

${\mathrm{T}}_2=13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}=-\frac{13.6}{9}\mathrm{e}\mathrm{V}$

${\mathrm{T}}_1:{\mathrm{T}}_2=\frac{1}{4}:\frac{1}{9}=9:4$

From $x-t$ graph,

$\begin{array}{rr}& A=1,T=8\\ & \Rightarrow \omega =\frac{2\pi }{T}\\ & \Rightarrow \omega =\frac{\pi }{4}\\ & \text{at}t=2,x=1\\ & a=-{\omega }^{2}x\\ & \Rightarrow \mathrm{a}=\frac{-{\pi }^2}{16}*1\\ & \Rightarrow \mathrm{a}=\frac{-{\pi }^2}{16}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Time period of satellite

$\mathrm{T}=2\pi \sqrt[]{\frac{{\mathrm{R}}^3}{\mathrm{G}\mathrm{M}}}$

$=2\pi \sqrt[]{\frac{{\mathrm{R}}^3}{\mathrm{G}\mathrm{d}\frac{4}{3}\pi {\mathrm{R}}^3}}$

$\Rightarrow \mathrm{T}=\sqrt[]{\frac{3\pi }{\mathrm{G}\mathrm{d}}}$ $\Rightarrow \frac{3\pi }{\mathrm{G}\mathrm{d}}={\mathrm{T}}^2$

$\frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{u}}{3}\right)}^2-\frac{1}{2}{\mathrm{m}\mathrm{u}}^2=-{\mathrm{F}}_{\mathrm{R}}*24$

$0-\frac{1}{2}{\mathrm{m}\mathrm{u}}^2=-{\mathrm{F}}_{\mathrm{R}}*\mathrm{d}$

$\frac{\frac{1}{2}{\mathrm{m}\mathrm{u}}^2}{\frac{1}{2}m{u}^2*\frac{8}{9}}=\frac{\mathrm{d}}{24}$

$\mathrm{d}=24*\frac{9}{8}=27\mathrm{c}\mathrm{m}$

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.