Physics MCQs for NEET — Practice Questions with Answers

Stopping potential= Maximum KE

                    eV=${\mathrm{KE}}_{\max }$

so, option (b) is correct.

Kinetic energy

                 KE=$\mathrm{\varphi }-{\mathrm{\varphi }}_0$

Here, $$\begin{gathered} {\mathrm{KE}}_1=1-0.5=0.5 \mathrm{eV} \\ {\mathrm{KE}}_2=2.5-0.5=2 \mathrm{eV} \end{gathered}$$

so,  $\frac{{\mathrm{KE}}_1}{{\mathrm{KE}}_2}=\frac{0.5}{2}=\frac{1}{4}$

or   $\frac{{\mathrm{v}}_1^2}{{\mathrm{v}}_2^2}=\frac{1}{4}$

or   $\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

By the concept of threshold minimum frequency needed for photoelectric emission.

  $$\begin{gathered} \frac{1}{2}m{v}^2=h(v\mathit{-}{v}_{\mathit{0}}) \\ \mathrm{v}\ge {\mathrm{v}}_0 \end{gathered}$$

We have

            $\mathrm{\lambda }=\frac{12.27}{\sqrt{\mathrm{V}}}$

So,        $$\begin{gathered} \frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2}=\sqrt{\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}} \\ {\mathrm{\lambda }}_2={\mathrm{\lambda }}_1\sqrt{\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}} \\ {\mathrm{\lambda }}_2={\mathrm{\lambda }}_1\sqrt{\frac{25}{100}} \\ {\mathrm{\lambda }}_2={\mathrm{\lambda }}_1\sqrt{\frac{1}{4}}=\frac{{\mathrm{\lambda }}_1}{2} \end{gathered}$$

Energy of incident light E(eV)=$\frac{12375}{2000}$

=6.2eV (200 nm= 2000$\stackrel{\circ }{A}$)

According to the relation E= $w_{\circ }$ + e$V_{\circ }$

$\Rightarrow$        $V_{\circ }$=$\frac{E-W_{\circ }}{e}$

                =$\frac{(6.2-5.01)e}{e}$

               =1.2 V

Stopping potential is negative, so answer is -1.2 V

Here $\lambda =667\times {10}^{-9}m, P=9\times {10}^{-3}W$

Power=$\frac{energy}{time}$

        =$\frac{nhc}{\lambda t}$

       =$\frac{Nhc}{\lambda }$

where N is number of photons emitted per sec.

$\Rightarrow N=\frac{P\times \lambda }{hc}$

            =$\frac{9\times {10}^{-3}\times 667\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8}$

           =$3\times {10}^{16}/s$

The number of photoelectrons emitted is directly proportional to the intensity of light.

According to laws of photoelectric effect

            $K{E}_{max}=E-\varphi$

where $\varphi$ is work function and $K{E}_{max}$ is maximum kinetic energy of photoelectron.

$\therefore hv=e{V}_0+\varphi$

or     $v=5 eV +6.2 eV =11.2eV$

$\therefore \lambda =\left(\frac{12400}{11.2}\right)\stackrel{0}{A}\approx 1000 \stackrel{0}{A}$

Hence, the radiation lies in ultraviolet region.

Wavelength of a particle is given by 

                  $\lambda =\frac{h}{p}$

where h is Planck's constant and wavelength of an electron is given by 

                         ${\lambda }_e=\frac{h}{p_e}$

but                 $\lambda ={\lambda }_e$

so,               p=$p_e$

or                mv=$m_e{v}_e$

or           $v=\frac{m_e{v}_e}{m}$

putting the under given data 

$$\begin{gathered} m_e=9.1\times {10}^{-31} kg , v_e=3\times {10}^6 m/s; \\ m=1 mg =1 \times {10}^{-6} kg \\ v=\frac{9.1\times {10}^{-31}\times 3\times {10}^6}{1\times {10}^{-6}} \\ =2.7\times {10}^{-18 } m{s}^{-1} \end{gathered}$$

(d) Light consists of photons and cathode rays consists of electrons. However both effect the photographic plate.