Question

The photoelectric threshold wavelength of a metal is 230 nm. Determine the maximum kinetic energy in joule and in eV of the ejects electron for the metal surface when it is exposed to a radiation of wavelength 180 nm.

[Planck’s constant : h = 6.63 * 10^-34 Js, Velocity of light : C = 3 * 10⁸ m/s.]

Answer 1

λ₀ = 230 nm = 230 × 10⁻⁹m,

λ = 180 nm = 180 × 10⁻⁹m,
h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸
m/s, K.E. (in joule and eV) = ?
Now,

∴ The maximum kinetic energy of the ejected electron is 2.402 × 10⁻¹⁹ J.

∴ The maximum kinetic energy of the ejected electron is 1.5 eV.