The threshold wavelength of silver is $3250 × 10^{-10}m$. The velocity of the electron ejected from the silver surface by ultra violet light of wavelength $2536 × 10^{-10}m$ is approximately:

(Given $h = 4.14 × 10^{-15} eVs$)

$6 × 10^5 ms^{-1}$

The correct answer is Option (1) → $6 × 10^5 ms^{-1}$

The energy of photon is given by -

$E_{photon}=\frac{hc}{λ}=\frac{(6.626×10^{-34})(3×10^8)}{2536×10^{-10}}$

$E_{work\,function}=\frac{hc}{λ_{threshold}}=\frac{(6.6×10^{-34})(3×10^8)}{3250×10^{-10}}$

$E_{kinetic}=E_{photon}-E_{work\,function}$

$=6.6×10^{-34}\left(\frac{3×10^8}{2536×10^{-10}}-\frac{3×10^8}{3250×10^{-10}}\right)$

and,

$E_{kinetic}=\frac{1}{2}mpv^2$

$⇒V=\sqrt{\frac{2E_{kinetic}}{mp}}$

$=\sqrt{\frac{2}{9.11×10^{-31}}×6.6×10^{-34}\left(\frac{3×10^8}{2536×10^{-10}}-\frac{3×10^8}{3250×10^{-10}}\right)}$

$≃6 × 10^5 m/s$