Question

A light of a 400nm is incident on a metal
with work function 3.1eV. The kinetic
energy of the photo electron emitted will
be
a) 6×10^9 J
b)6×10^19 ergs
c)zero
d)3×10^19J​

Answer 1

Given:

A light of a 400nm is incident on a metal with work function 3.1eV.

To find:

KE of emitted electrons?

Calculation:

Let's find out the energy of incident photons:

$\sf \: E =  \dfrac{hc}{ \lambda}$

$\sf  \implies\: E =  \dfrac{6.63 \times  {10}^{ - 34}  \times (3 \times  {10}^{8} )}{400 \times  {10}^{ - 9} }$

$\sf  \implies\: E =  \dfrac{19.89 \times  {10}^{ - 26} }{4 \times  {10}^{ - 7} }$

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

$\sf  \implies\: E =  \dfrac{4.97 \times  {10}^{ - 19} }{e} \: eV$

$\sf  \implies\: E =  \dfrac{4.97 \times  {10}^{ - 19} }{1.6 \times  {10}^{ - 19} } \: eV$

$\sf  \implies\: E =  3.1 \: eV$

Since energy of incident photon is equal to the work function of the metal, KE of electrons will be zero.

$\sf KE = E -  w_{0}$

$\sf  \implies KE = 3.1 - 3.1$

$\sf  \implies KE =0 \: joule$

Hope It Helps