Physics MCQs for NEET — Practice Questions with Answers

(d) 

$$\begin{gathered} \mathrm{hv}-{\mathrm{W}}_0=\frac{1}{2}{\mathrm{mv}}_{\max }^2\Rightarrow \frac{\mathrm{hc}}{\mathrm{\lambda }}-\frac{\mathrm{hc}}{{\mathrm{\lambda }}_0}=\frac{1}{2}{\mathrm{mv}}_{\max }^2 \\ \Rightarrow \mathrm{hc}\left(\frac{{\mathrm{\lambda }}_0-\mathrm{\lambda }}{{\mathrm{\lambda \lambda }}_0}\right)=\frac{1}{2}{\mathrm{mv}}_{\max }^2\Rightarrow {\mathrm{v}}_{\max }=\sqrt{\frac{2\mathrm{hc}}{\mathrm{m}}\left(\frac{{\mathrm{\lambda }}_0-\mathrm{\lambda }}{{\mathrm{\lambda \lambda }}_0}\right)} \end{gathered}$$    

When wavelength is $\mathrm{\lambda }$ and velocity is v, then 

$\mathrm{v}=\sqrt{\frac{2\mathrm{hc}}{\mathrm{m}}\left(\frac{{\mathrm{\lambda }}_0-\mathrm{\lambda }}{{\mathrm{\lambda }}_0\mathrm{\lambda }}\right)}$               .....(i) 

When wavelength is $\frac{3\mathrm{\lambda }}{4}$ and velocity is v', then 

$\mathrm{v}\text{'}=\sqrt{\frac{2\mathrm{hc}}{\mathrm{m}}\left(\frac{{\mathrm{\lambda }}_0-3\mathrm{\lambda }/4}{{\mathrm{\lambda }}_0\times \left(3\mathrm{\lambda }/4\right)}\right)}$      .....(ii) 

Divide equation (ii) by (i), we get

$$\begin{gathered} \frac{\mathrm{v}\text{'}}{\mathrm{v}}=\sqrt{\left(\frac{\left[{\mathrm{\lambda }}_0-\left(3\mathrm{\lambda }/4\right)\right]}{{\displaystyle \frac{3}{4}}{\mathrm{\lambda \lambda }}_0}\right)\times \frac{{\mathrm{\lambda \lambda }}_0}{{\mathrm{\lambda }}_0-\mathrm{\lambda }}} \\ \mathrm{v}\text{'}=\mathrm{v}{\left(\frac{4}{3}\right)}^{1/2}\sqrt{\frac{\left[{\mathrm{\lambda }}_0-\left(3\mathrm{\lambda }/4\right)\right]}{{\mathrm{\lambda }}_0-\mathrm{\lambda }}} \mathrm{i}.\mathrm{e}. \mathrm{v}\text{'}>\mathrm{v}{\left(\frac{4}{3}\right)}^{1/2} \end{gathered}$$

(b) By using $\mathrm{hv}-{\mathrm{hv}}_0={\mathrm{K}}_{\max }$

$\Rightarrow \mathrm{h}\left({\mathrm{v}}_1-{\mathrm{v}}_0\right)={\mathrm{K}}_1$                        .....(i) 

And $\Rightarrow \mathrm{h}\left({\mathrm{v}}_2-{\mathrm{v}}_0\right)={\mathrm{K}}_2$                 .....(ii)

$\Rightarrow \frac{{\mathrm{v}}_1-{\mathrm{v}}_0}{{\mathrm{v}}_2-{\mathrm{v}}_0}=\frac{{\mathrm{K}}_1}{{\mathrm{K}}_2}=\frac{1}{\mathrm{K}}, \mathrm{Hence} {\mathrm{v}}_0=\frac{{\mathrm{Kv}}_1-{\mathrm{v}}_2}{\mathrm{K}-1}$

(c) de-Broglie wavelength $\mathrm{\lambda }=\frac{\mathrm{h}}{{\mathrm{mv}}_{\mathrm{rms}}}$, rms velocity of a gas particle at the given temperature (T) is given as

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}_{\mathrm{rms}}^2=\frac{3}{2}\mathrm{kT}\Rightarrow {\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{kT}}{\mathrm{m}}}\Rightarrow {\mathrm{mv}}_{\mathrm{rms}}=\sqrt{3\mathrm{mkT}} \\ \therefore \mathrm{\lambda }=\frac{\mathrm{h}}{{\mathrm{mv}}_{\mathrm{rms}}}=\frac{\mathrm{h}}{\sqrt{3\mathrm{mkT}}} \\ \Rightarrow \frac{{\mathrm{\lambda }}_{\mathrm{H}}}{{\mathrm{\lambda }}_{\mathrm{He}}}=\sqrt{\frac{{\mathrm{m}}_{\mathrm{He}}{\mathrm{T}}_{\mathrm{He}}}{{\mathrm{m}}_{\mathrm{H}}{\mathrm{T}}_{\mathrm{H}}}}=\sqrt{\frac{4\left(273+127\right)}{2\left(273+27\right)}}=\sqrt{\frac{8}{3}} \end{gathered}$$

(b) The momentum of the incident radiation is given as $\mathrm{p}=\frac{\mathrm{h}}{\mathrm{\lambda }}$. When the light is totally reflected normal to the surface the direction of the ray is reversed. That means it reverses the direction of it’s momentum without changing it’s magnitude
$\therefore ∆\mathrm{p}=2\mathrm{p}=\frac{2\mathrm{h}}{\mathrm{\lambda }}=\frac{2\times 6.6\times {10}^{-34}}{6630\times {10}^{-10}}=2\times {10}^{-27 }$ kg-m/sec.

(c) When a charged particle (charge q, mass m) enters perpendicularly in a magnetic field (B) than, radius of the path described by it

$\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}\Rightarrow \mathrm{mv}=\mathrm{qBr}$

Also the de-Broglie wavelength $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$

$\Rightarrow \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{qBr}}\Rightarrow \frac{{\mathrm{\lambda }}_{\mathrm{\alpha }}}{{\mathrm{\lambda }}_{\mathrm{P}}}=\frac{{\mathrm{q}}_{\mathrm{P}}{\mathrm{r}}_{\mathrm{P}}}{{\mathrm{q}}_{\mathrm{\alpha }}{\mathrm{r}}_{\mathrm{\alpha }}}=\frac{1}{2}$

(b)

$Cutoff wavelength (in \stackrel{o}{A})= \frac{12375}{\varphi (in eV)} = 2578 \stackrel{o}{A}$

Photoelectric emission is possible with ${\mathrm{\lambda }}_1$ only. Maximum kinetic energy of emitted photoelectrons K = E – ${\mathrm{W}}_0$ = 5 – 4.8 = 0.2 eV.
Photo electrons experience magnetic force and move along a circular path of radius

$\mathrm{r}=\frac{\sqrt{2\mathrm{mk}}}{\mathrm{QB}}=\frac{\sqrt{2\times 9\times {10}^{-31}\times 0.2\times 1.6\times {10}^{-19}}}{1.6\times {10}^{-19}\times 3\times {10}^{-5}}$

= 0.05 m = 5 cm.

(a) Momentum of a photon is given by $\mathrm{p}=\frac{\mathrm{h}}{\mathrm{\lambda }}$
Also the photon is a form of energy packets behaves as a particle having energy $\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$. So $\mathrm{p}=\frac{\mathrm{E}}{\mathrm{c}}$

(d) Photoelectric effect demonstrates the particle nature of light. Number of emitted photoelectrons depends upon the intensity of light.

(b) Charge does not change with speed but mass varies with the speed as per relation $\mathrm{m}=\frac{{\mathrm{m}}_0}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}}$. Hence specific charge e/m decreases with increase in speed.

(b) Mass of moving photon $\mathrm{m}=\frac{\mathrm{h\nu }}{{\mathrm{c}}^2}=\frac{\mathrm{h}}{\mathrm{c\lambda }}$ and $\mathrm{E}={\mathrm{mc}}^2$.