Question

Light of frequency 7.21 x 10^4 Hz is incident on a metal surface. The cut-off wavelength for
photoelectric emission from the metal surface is 540 nm. Determine the maxim n speed of the
photoelectrons emitted from the surface. (Given: h= 6.63 x 10-34 Js, mass of an electron = 9.1 x 10-31
kg).​

Answer 1

Answer:

Explanation:

Given frequency of light= 7.21*10^4 Hz

cut off wavelength= 540 nm

KE(max)= 1/2mv^2= h(v-v0)

v0= v- MV^2/2h

V0= 7.21- 9.1*10^-31 * (3*10^8)^2/2(6.63*10^-34)

V0= 4.74* 10^14 Hz

Note: in the given question, speed should be given , and we need to find threshold frequency.

Answer 2

Answer:

Explanation: Maximum speed