Question

When light with a wavelength of 247 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.84 × 10^-19 J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answer 1

When light with a wavelength of 247 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.84 10-19 J. Determine the wavelength of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Energy of light = h * f
f * wave length = c
f = c ÷ wave length
E = h * c ÷ λ
h = 6.67 * 10^-34
c =3.0 * 10^8
λ = 247 nm = 247 * 10^-9 = 2.47 * 10^-7

Energy of 247 nm light = h * c ÷ λ = 6.67 * 10^-34 * 3.0 * 10^8 ÷ (2.47 * 10^-7) = 8.101 * 10^-19 J

This is the amount of energy which the light gave to the atom.
The maximum kinetic energy of electron = 2.84 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.
8.101 * 10^-19 – 2.84 * 10-19 = 5.26 * 10^19 Joules of energy was absorbed by the rest of the atom

Total kinetic energy of electron with new wave length = 2 * 2.84 10-19 = 5.68 * 10^-19 J

Add the energy absorbed by the rest of the atom
5.68 * 10^-19 + 5.26 * 10^-19 = 1.094 * 10^-18 J
This is the energy of light at the new wave length

1.094 * 10^-18 = 6.67 * 10^-34 * 3.0 * 10^8 ÷ λ
λ = 1.829 * 10^-7 m = 182.91 nm