Question

The minimum energy required to overcome the nuclear attractive forces of Ag atom is 5.52 x 10-19 J. What

will be kinetic energy of ejected electrons if the incident radiation is of λ=360 Å. Use h=6.626 x 10-34 J-s)​

Answer 1

Given:

The minimum energy required to overcome the nuclear attractive forces of Ag atom is 5.52 x 10-19 J.

To find:

Kinetic energy of ejected electrons?

Calculation:

So, we will apply EINSTEIN'S PHOTOELECTRIC EFFECT EQUATION as follows:

$\sf \: KE = \dfrac{hc}{ \lambda} - w_{0}$

$\sf \implies\: KE = \dfrac{6.6 \times {10}^{ - 34} \times (3 \times {10}^{8} )}{ 360 \times {10}^{ - 10} } - (5.52 \times {10}^{ - 19} )$

$\sf \implies\: KE = \dfrac{19.8 \times {10}^{ - 26}}{ 3.6 \times {10}^{ - 8} } - (5.52 \times {10}^{ - 19} )$

$\sf \implies\: KE = ( 5.5 \times {10}^{ - 18} ) - (5.52 \times {10}^{ - 19} )$

$\sf \implies\: KE = ( 55 \times {10}^{ - 19} ) - (5.52 \times {10}^{ - 19} )$

$\sf \implies\: KE \approx \: 49.5 \: \times {10}^{ - 19} \: joule$

So, kinetic energy of electrons is 49.5 × 10^(-19) Joule.