Question

In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper (φ = 4.5 eV). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector.

Answer 1

Explanation:

To Find: The minimum and maximum kinetic energy of the photoelectrons reaching the collector.

Step 1:

Given data in the question,  

Work function, ϕ = 4.5 eV,

Monochromatic Wavelength of light, λ = 200 nm

From the photoelectric equation of Einstein,

Kinetic energy,

$K=E-\phi=\frac{h c}{\lambda}-\phi$

where, h = Planck's constant          

           c = speed of light  

Step 2:

On substituting the values we get,

$\therefore K=\frac{1242}{200}-4.5$

       = 6.21 - 4.5

       = 1.71 eV

Therefore it requires at least 1.7 eV to avoid the electron. Maximum kinetic energy is therefore to be 2 eV.  

It is provided that the electron requires 2 V of electrical potential to accelerate. Hence full kinetic energy

= (2+ 1.7) eV = 3.7 eV