Question

When light with a wavelength of 215 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.33 × 10 − 19 J. Determine the wavelength of light that should be used to quadruple the maximum kinetic energy of the electrons ejected from this surface.

Answer 1

Expression for the energy of light wave is given as -

E = h * f

Also we have -

f * wave length = c

f = c ÷ wave length
E = h * c ÷ λ ----------------------------------------------------(i)
h = 6.67 * 10^-34
c =3.0 * 10^8 m/s

λ = 215 nm = 215 * 10^-9 = 2.15 * 10^-7 m

Energy of 215 nm light = h * c ÷ λ = (6.67 * 10^-34 * 3.0 * 10^8) / (2.38 * 10^-7) = 8.40 * 10^-19 J

This is the amount of energy which the light gave to the atom.
The maximum kinetic energy of electron = 3.33 10-19 J
This is the energy which 1 electron absorbed.
The difference between these 2 energies is the amount of energy absorbed by the atom.

Means -

8.40 * 10^-19 – 3.33 * 10-19 = 5.07 * 10^-19 Joules of energy was absorbed by the rest of the atom

As mentioned in the problem, total kinetic energy of electron with new wave length = 4 * 3.33 * 10-19 = 13.32 * 10^-19 J

Add the energy absorbed by the rest of the atom
13.32 * 10^-19 + 5.07 * 10^-19 = 1.839 * 10^-18 J

This is the energy of light at the new wave length

Put this value in expression (i) to determine the wavelength of light -

1.839 * 10^-18 = (6.67 * 10^-34 * 3.0 * 10^8) / λ

=> λ = 1.088 * 10^-7 m = 108.8 nm (Answer)