Question

Light of frequency 7.21 × 1014 Hz is incident on a metal surface. The cut-off wavelength for photoelectric emission from the metal surface is 540 nm. Determine the maximum speed of the photoelectrons emitted from the surface.
(Given: h = 6.63 × 10-34 Js, mass of an electron = 9.1 × 10-31 kg).

Answer 1

Given:

Light of frequency 7.21 × 1014 Hz is incident on a metal surface. The cut-off wavelength for photoelectric emission from the metal surface is 540 nm.

To find:

Max speed of photo-electrons ?

Calculation:

According to Einstein's Photo-electric Effect equation:

$\therefore \: KE = h \nu - W$

$\implies \: \dfrac{1}{2}m {v}^{2} = h \nu - \dfrac{hc}{ \lambda}$

$\implies \: \dfrac{1}{2}m {v}^{2} = (6.63 \times {10}^{ - 34} \times 7.21 \times {10}^{14}) - \dfrac{6.63 \times {10}^{ - 34} \times 3 \times {10}^{8} }{540 \times {10}^{ - 9} }$

$\implies \: \dfrac{1}{2}m {v}^{2} = (47.8 \times {10}^{ - 20}) - \dfrac{6.63 \times {10}^{ - 34} \times 3 \times {10}^{8} }{5.4 \times {10}^{ - 7} }$

$\implies \: \dfrac{1}{2}m {v}^{2} = (47.8 \times {10}^{ - 20}) - (36.8 \times {10}^{ - 20} )$

$\implies \: \dfrac{1}{2}m {v}^{2} = 11 \times {10}^{ - 20}$

$\implies \: m {v}^{2} = 22 \times {10}^{ - 20}$

$\implies \: {v}^{2} = \dfrac{ 22 \times {10}^{ - 20}}{9.1 \times {10}^{ - 31} }$

$\implies \: {v}^{2} = \dfrac{ 22 \times {10}^{ - 20}}{0.91 \times {10}^{ - 30} }$

$\implies \: {v}^{2} = 24.17 \times {10}^{ 10}$

$\implies \: v =4.9 \times {10}^{5} \: m {s}^{ - 1}$

So, velocity of photo-electrons is 4.9 × 10⁵ m/s.