Physics MCQs for NEET — Practice Questions with Answers

(a) Key idea

Cut-off wavelength occurs when an incoming electron loses its complete energy in the collision. This energy appears in the form of X-rays.

Given, the mass of electrons=m

    de-Broglie wavelength=$\lambda$

So, kinetic energy, of electron = $\frac{p^2}{2m}$

                                        $=\frac{1}{2m}{\left(\frac{h}{\lambda }\right)}^2=\frac{h^2}{2m{\lambda }^2}$

Now, the maximum energy of a photon can be given by-

$$\begin{gathered} E=\frac{hc}{{\lambda }_0}=\frac{h^2}{2m{\lambda }^2} \\ \Rightarrow {\lambda }_0=\frac{hc\times 2{\lambda }^{2}m}{h^2} \\ = \frac{2mc{\lambda }^2}{h^{}} \end{gathered}$$

(c) In Ist case, when a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V.

So, photoelectric equation can be written as

              eV=$\frac{hc}{\lambda }-\frac{hc}{{\lambda }_o}$                        ...(i)

In IInd Case, when the same surface is illuminated with radition of wavelength 2$\lambda$, the stopping potential is $\frac{V}{4}$ So, photoelectric equation can be written as

                        $\frac{eV}{4}=\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_0}$

$\Rightarrow eV=\frac{4hc}{2\lambda }-\frac{4hc}{{\lambda }_0}$                ...(ii)

From eqs, (i) and (ii) we get

$$\begin{gathered} \Rightarrow \frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=\frac{4hc}{2\lambda }-\frac{4hc}{{\lambda }_o} \\ \Rightarrow \frac{1}{\lambda }-\frac{1}{{\lambda }_o}= \frac{2}{\lambda }- \frac{4}{{\lambda }_o} \\ \Rightarrow {\lambda }_0=3\lambda \end{gathered}$$

(c)

As energy of photon,E=hv
E=hc/λ
$E (in eV) =\frac{12375}{\lambda (in {\displaystyle \stackrel{o}{A}})} = 2.48 eV$

According to Einstein's photoelectric emission,we have

KE_max=E-W=2.48-2.28=0.2eV

For de-Broglie wavelength of the emitted electron.

λ_{e min}=12.27/√KE_max(eV)

=12.27/√0.2

=27.436Å

=27.436x10^-10 m

(b)

$$\begin{gathered} Energy flux=25\times {10}^4 J/s-m^2 \\ Force on unit area=Momentum transferred in unit time on area=2\frac{h}{\lambda }=2\frac{E}{c}=\frac{2\times 25\times {10}^4 J/s-m^2}{3\times {10}^8} \\ Force on the total area=\frac{2\times 25\times {10}^4 J/s-m^2}{3\times {10}^8}\times 15\times {10}^{-4}=2.5\times {10}^{-6}N \end{gathered}$$

(a) 

$$\begin{gathered} Wavelength of the electron: \\ {\lambda }_e=\frac{h}{\sqrt{2mE}} \\ Wavelegth of photon: \\ {\lambda }_p=\frac{hc}{E} \\ \Rightarrow {{\lambda }_e}^2=\frac{h^2}{2mE}=\frac{h^2{\lambda }_p}{2mhc} \\ \Rightarrow {\lambda }_p\propto {{\lambda }_e}^2 \end{gathered}$$

Efficient power P =$\frac{N}{t}\times \frac{hc}{\lambda }=200\times 0.25$

                      $\frac{N}{t}=50\times \frac{\lambda }{hc}=1.5\times {10}^{20}$

                           $= \frac{50\times 0.6\times {10}^{-6}}{6.6\times {10}^{-34}\times 3\times {10}^8}$

We knows

     $R=\frac{mv}{qB}$

and $\lambda =\frac{h}{mv}$

     $\lambda =\frac{h}{qBR}$

       $=\frac{6.6\times {10}^{-34}}{2\times 1.6\times {10}^{-19}\times 0.83\times {10}^{-2}\times 0.25}$

       $=0.01 \stackrel{0}{A}$

de-Broglie wavelength

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

Here       $\left|\frac{∆\mathrm{\lambda }}{\mathrm{\lambda }}\right|=\left|\frac{∆\mathrm{p}}{\mathrm{p}}\right|$

             $$\begin{gathered} \frac{0.5}{100}=\frac{\mathrm{P}}{{\mathrm{p}}_{\mathrm{i}}} \\ {\mathrm{P}}_{\mathrm{i}}=\frac{100}{0.5}\mathrm{p} \\ {\mathrm{P}}_{\mathrm{i}}=200\mathrm{p} \end{gathered}$$

In the Davisson and Germer experiment, the 

velocity of the electron emitted from the electron 

gun can be increased by increasing the potential

difference between the anode and filament.

Momentum of photon

          $\mathrm{p}=\frac{\mathrm{h\nu }}{\mathrm{c}}$

Hence, Recoil energy

        $$\begin{gathered} \mathrm{E}=\frac{{\mathrm{P}}^2}{2\mathrm{M}} \\ \mathrm{E}=\frac{{\left({\displaystyle \frac{\mathrm{h\nu }}{\mathrm{c}}}\right)}^2}{2\mathrm{M}} \end{gathered}$$

or            $h=\frac{h^2{\nu }^2}{2M{c}^2}$