Question

A photon of wavelength 4000 A strikes a metal
surface having work function of 1.9 x 10-19 J. The
kinetic energy of the emitted photoelectron is
3.05 x 10-19
4.25 x 10-19
2.20 x 10-19
No electron will be emitted

Answer 1

Given:

A photon of wavelength 4000 A strikes a metal

surface having work function of 1.9 x 10-19 J.

To find:

Kinetic Energy of the emitted electrons.

Calculation:

Energy of photons:

$\therefore \: E = \dfrac{hc}{ \lambda}$

$= > E = \dfrac{6.634 \times {10}^{ - 34} \times 3 \times {10}^{8} }{4000 \times {10}^{ - 10} }$

$= > E = \dfrac{6.634 \times {10}^{ - 34} \times 3 \times {10}^{8} }{4 \times {10}^{ - 7} }$

$= > E = \dfrac{6.634 \times {10}^{ - 34} \times 3 \times {10}^{15} }{4 }$

$= > E = 4.97 \times {10}^{ - 19} \: joule$

So, energy of the emitted electrons is

$KE = E -( 1.9 \times {10}^{ - 19} )$

$= > KE = (4.97 \times {10}^{ - 19} )-( 1.9 \times {10}^{ - 19} )$

$= > KE = 3.07 \times {10}^{ - 19}$

$= > KE \approx 3.05 \times {10}^{ - 19} \: joule$

So, final answer is:

$\boxed{ \sf{KE \approx 3.05 \times {10}^{ - 19} \: joule}}$

Answer 2

Explanation:

I hope it helps you......