Question

A photosensitive surface of work function 2.1 eV is irradiated by radiation of wavelength 150 nm. Calculate

1. the threshold wavelength, 
2. energy (in eV) of an incident photon, and 
3. maximum kinetic energy of emitted photoelectron.

Answer 1

i. Here, Φ₀ = 2.1 eV

= 2.0 × 1.6 × 10⁻¹⁹ J

λ = 150 × 10⁻⁹ m

As, `phi_0 = (hc)/(lambda_0)`

or `lambda_0 = (hc)/(phi_0)`

= `((6.63 xx 10^-34) xx (3 xx 10^8))/(2.1 xx 1.6 xx 10^-19)`

= 6.187 × 10⁻⁷ m

= 6187.5 Å

ii. Max. kinetic energy of photoelectron is

`K_(max) = (hc)/lambda - phi_0`

= `((6.63 xx 10^-34) xx (3 xx 10^8))/((150 xx 10^-9) xx (1.6 xx 10^-19)) -2.0` 

= 7.42 − 2.1

= 5.32 eV

iii. Stopping potential `V_0 = (K_(max))/e`

= `(5.32 eV)/e`

= 5.32 V