Question

A light source of wavelength \lambdaλ illuminates a metal with a work function of \text{BE} = 2.00 ~\text{eV}BE=2.00 eV and ejects electrons with a maximum ~\text{KE} = 4.00 ~\text{eV} KE=4.00 eV. A second light source with double the wavelength of the first ejects photoelectrons with what a maximum kinetic energy?

Answer 1

Answer:

1 eV

Explanation:

Given:

Work function, ∅ = 2.00 eV

Kinetic energy of the ejected of the electron, K.E = 4.0 eV

Now,

using the photoelectric equation , we have

Energy of the photon (E) = ∅ + K.E

also,

E = hc/λ

where, h is plank's constant

c is the speed of the light

λ is the wavelength

thus, we have

hc/λ = 2 + 4 = 6 eV

Energy of photon = 6eV

Now,

for the second case

λ' = 2λ

when Wavelength is doubled , E is halved

thus,

E' = hc/λ'

or

E' = hc/2λ

or

E' = E/2 = 6/2 = 3 eV

also,

E' = ∅ + KE '

thus on substituting the values,

3 = 2 + KE'

or

KE' = 1 eV

Hence, the maximum kinetic energy for the second case is 1 eV