Question

a light beam is inciden on a metal target with a work function 2.3 eV.if a stopping potential of 1.3V is required, what is the wavelength of the incident light

Answer 1

The relationship between stopping potential and kinetic energy :

K. E  = Stopping potential × electric charge.

Electric charge = 1.6022 × 10⁻¹⁹

Our stopping potential = 1.3eV

K. E = 1.6022 × 10⁻¹⁹ × 1.3 = 2.087 × 10⁻¹⁹

The relationship between Kinetic energy and work function :

K. E = hf - Ф

Where :

h = planks constant = 6.62 × 10⁻³⁴

f = frequency

Ф = Work function.

Doing our substitution we have :

2.08727 × 10⁻¹⁹ = hf - 2.3

hf = 2.0827 × 10⁻¹⁹ + 2.3

f = (2.0827 × 10⁻¹⁹ + 2.3) / (6.62 × 10⁻³⁴)

f = 3.474 × 10⁻³³

The relationship between, velocity, wavelength and frequency of light.

f = c/y

Where:

c = speed of light = 3 × 10⁸

y = wavelength.

Doing substitution :

3.474 × 10⁻³³ = (3.0 × 10⁸) /y

y = (3.0 × 10⁸) /(3.474 × 10⁻³³) = 8.636 × 10⁴⁰m

Answer 2

K.E = stopping potential × electric charge

= 1.3×1.6×10^(-19) J

= 2.08×10(-19) J

Also KE = E - work function

Here , E= h×c÷y

h = 6.63 × 10^(-34) J s

c= 3× 10^(8) m/s

y is wavelength which we have to find out

Work function = 2.3eV

I eV = 16 × 10^(-19) J

So , work function = 2.3×1.6×10^(-19)

= 3.68×10^(-19) J

Using equation ,

KE = E - work function

E = 5.76×10(-19) J

Also E = h×c÷y

y= 3.45×10^(-7) m