Question

7. The threshold wavelength for a photoelectric surface is 6000A and the wavelength of incident light is
5000A. Then the maximum energy of emitted electrons would be:
a. 0.41 eV
b.0.041eV
c.4.1eV
d.41eV​

Answer 1

Answer:

a) 0.41ev

Explanation:

On using Einstein equation for photoelectric emission

$k(max) = \frac{hc}{frequncy \: of \: incident \: radiation} - \frac{hc}{threshold \: frequency}$

On putting values in the equation:-

$k(max) = \frac{20 \times {10}^{ - 26} }{5 \times {10}^{ - 7} } \times \frac{20 \times {10}^{ - 26} }{6 \times {10}^{ - 7} }$

$k(max) = 20 \times {10}^{ - 19} \times ( \frac{1}{5} - \frac{1}{6} )$

$k(max) = \frac{20 \times {10}^{ - 19} }{30 \times 1.6 \times {10}^{ - 19} }$

1.6*10^-19 is divided for converting energy from joule to ev

$k(max) = 0.41ev$