Question

calculate the maximum kinetic energy of photoelectrons emitted when a light of frequency 2 into 10 to the power 16 hertz is irradiated on a metal surface with threshold frequency V• equal to 8.68 into 10 to the power 15 Hertz ​

Answer 1

Answer:

The maximum kinetic energy of photo-electrons is 7.505 * 10^(-20) J.

Explanation:

Given data:

Absorbed frequency of the photo-electrons, f = 2 * 10¹⁶ Hz  

Threshold frequency of the metal surface, fo = 8.68 * 10¹⁵ Hz

To find: Maximum K.E.

We have,  

Maximum K.E. = h(Absorbed frequency – Threshold frequency)

K.E. = h ( f – fo ) …. [Plank’s constant h=6.63 * 10⁻³⁴Js]

Or, K.E. = 6.63 * 10⁻³⁴ * [{2*10¹⁶} – {8.68*10¹⁵}]

Or, K.E. = 6.63 * 10⁻³⁴ * 10¹⁵ * [{2*10} – 8.68]

Or, K.E. = 6.63*10⁻¹⁹ * 11.32

Or, K.E. = 75.0516 *10⁻¹⁹

Or, K.E. = 7.505 * 10⁻²⁰ J