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

3.32×10⁻²⁰ J

Step-by-step 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) = (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) = (6.626* 10^(-34)* 3* 10^(8))/(310* 10^(-9) ) - 3.8* 1.6* 10^(-19)`

`K.E_(max) = 3.32* 10^(-20)J`