Question

Radiation of wavelength 200 nm is incident on a photosensitive surface of work function 4.2 eV. The kinetic energy of fastest photoelectrons emitted from this surface will be close to ______.

Options

• 3.5 eV

• 3.0 eV

• 2.5 eV

• 2.0 eV

Answer 1

Radiation of wavelength 200 nm is incident on a photosensitive surface of work function 4.2 eV. The kinetic energy of fastest photoelectrons emitted from this surface will be close to 2.0 eV.

Explanation:

Using Einstein’s photoelectric equation:

`K_max = (hc)/λ - ϕ`

hc = 3 × 10⁸ × 6.63 × 10⁻³⁴ 

= 19.89 × 10⁻²⁶ J

= 1.989 × 10⁻²⁵ J

Convert Joule into eV

= `(1.989 xx 10^-25)/(1.6 xx 10^-19)`

= 1.24 × 10⁻⁶

Convert meters into nanometers

= 1.24 × 10⁻⁶ × 10⁹   

= 1.24 × 10³

= 1240 eV

E(eV)  = `(hc)/(λ(nm))`

= `1240/200`

= 6.2

∴ K_max = E − ϕ

= 6.2 − 4.2

= 2.0 eV