Question

Light with a wavelength of 310 nm is incident on a metal that has a work function of 3.8 eV. What is the maximum kinetic energy that a photoelectron ejected in this process can have?

Answer 1

Answer:

3.32×10⁻²⁰ J

Explanation:

Given :

Wavelength of the light λ = 310 nm

work function, W₀ = 3.8 eV

The maximum kinetic energy ($K.E_{max}$) is given as:

$K.E_{max} = \frac{hc}{\lambda } - W_{0}$

Where,

h = Planck's constant (6.626 × 10⁻³⁴ Js)

c = speed of light (3 × 10⁸)

also, 1 eV = 1.6 × 10⁻¹⁹ J

substituting the values in the above equation we get

$K.E_{max} = \frac{6.626\times 10^{-34}\times 3\times 10^{8}}{310\times 10^{-9} } - 3.8\times 1.6\times 10^{-19}$

$K.E_{max} = 3.32\times 10^{-20}J$