Question

When light of frequency 1.30x 1015 s-1 shines on the surface of cesium metal, electrons are ejected with a maximum kinetic energy of 5.2x 10-19 J. Calculate the wavelength of this light in nm.

Answer 1

2307.7 Angstrom

Step-by-step explanation:

frequency, f = 1.30 x 10^15 per second

Planks's constant, h = 6.63 x 10^-34 Js

velocity of light, c = 3 x 10^8 m/s

Let λ be the wavelength.

λ = c / f

λ = ( 3 x 10^8) / (1.30 x 10^15) = 2.308 x 10^-7 m

λ = 2307.7 Angstrom

Thus, the wavelength is 2307.7 Angstrom.