Calculate the
(a)    momentum, and
(b)    de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V. 

Given,
Frequency of light, $\mathrm{\nu }$ = 7.21 x 10¹⁴ Hz
Maximum speed of electrons, v_max = 6.0 x 10⁵ ms^–1
Mass of the electron, m = 9 x 10^–31 kg

Applying Einstein's photoelectric equation,
                $\mathrm{K}.\mathrm{E}{.}_{\max } = \frac{1}{2}{{\mathrm{mv}}^2}_{\max } = \mathrm{h}(\mathrm{v}-{\mathrm{v}}_0)$

$\Rightarrow$       $\boxed{\frac{1}{2}{{\mathrm{mv}}^2}_{\max } = \mathrm{h}(\mathrm{v}-{\mathrm{v}}_0)}$
$\Rightarrow$                  ${\mathrm{v}}_0 = \mathrm{v} - \frac{{{\mathrm{mv}}^2}_{\max }}{2\mathrm{h}}$
$\Rightarrow$                              ${\mathrm{v}}_0 = 7.21 \times {10}^{14} - \frac{(9.1 \times {10}^{-31}) \times (6 \times {10}^5{)}^2}{2 \times (6.63 \times {10}^{-34})}$
                           $= 4.74 \times {10}^{14}\mathrm{Hz}.$

given, 
Kinetic energy of the electron, K.E = 120 eV

(a) Momentum of the electron is given as, $\mathrm{p} = \sqrt{2\mathrm{mE}}$ 

$\Rightarrow$ $$\begin{gathered} \mathrm{p} = \sqrt{2 \times (9 \times {10}^{-31}) \times (120 \times 1.6 \times {10}^{-19})} \\ = 5.88 \times {10}^{-24} \mathrm{kg} {\mathrm{ms}}^{-1} \end{gathered}$$

(b) Now, since $\mathrm{p} = \mathrm{mv}$ 
we have,      
                     $$\begin{gathered} v = \frac{\mathrm{p}}{\mathrm{m}} \\ = \frac{5.88 \times {10}^{-24}}{9.1 \times {10}^{-31}} \end{gathered}$$ 
                              
$\Rightarrow$                 $\mathrm{v} = 6.46 \times {10}^6 \mathrm{m}/\mathrm{s}$ 

(c) De- broglie wavelength of electron is given by, 
                 $$\begin{gathered} \boxed{\mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}}} \\ = \frac{6.63 \times {10}^{-34}}{5.88 \times {10}^{-24}} \\ = 1.13 \times {10}^{-10}\mathrm{m} \end{gathered}$$

                    $= 1.13 \mathrm{\AA }.$                

Given,
Wavelength of light, λ = 488 nm = 488 x 10^–9 m
Stopping potential, V₀ = 0.38V 

As,            $\boxed{\boxed{{\mathrm{eV}}_0 = \mathrm{hv} - {\mathrm{\varphi }}_0}}$

$\Rightarrow$             ${\mathrm{eV}}_0 = \mathrm{h}\frac{\mathrm{c}}{\mathrm{\lambda }}-{\mathrm{\varphi }}_0$

$\Rightarrow$              ${\mathrm{\varphi }}_0 = \frac{\mathrm{hc}}{\mathrm{\lambda }}-{\mathrm{eV}}_0$

                           $= \left(\frac{6.63 \times {10}^{-34} \times 3 \times {10}^8}{488 \times {10}^{-9} \times 1.6 \times {10}^{-19}}-\frac{1.6 \times {10}^{-19} \times 0.38}{1.6 \times {10}^{-19}}\right)$

                      $$\begin{gathered} = (2.55 - 0.38) \mathrm{eV} \\ = 2.17 \mathrm{eV}. \end{gathered}$$