Question

Relation between threshold frequency, frequency of the incident photon and cutoff potential​

Answer 1

Answer:

hv' = hv - eV

Where, h = Planck's constant

v' = threshold frequency

v = frequency of incident

photon

e = charge of 1 electron

V = stopping potential

Explanation:

Using Einstein's photoelectric equation

hv = hv' + KE(max)

To terminate the photoelectric current stopping potential (V) is needed to apply in the opposite direction of the photoelectric current.

Work done by the stopping potential is equal to the maximum kinetic energy of the photoelectrons

Hence,

KE(max) =( work done by

stopping potential )

= (charge) * (potential)

KE(max) = eV

Substituting in photoelectric equation

hv = hv' + eV

This is the required relation

Answer 2

$eV=$$h$ν-$h$ν', is the relation between threshold frequency, frequency of the incident photon and the cut off potential.

Threshold frequency(ν')

It is the minimum frequency required for the photoelectric effect to take place.

Frequency of incident photon(ν)

The frequency at which the photon will be incident on the metal surface.

Cut off potential(V)

It is also called as stopping potential and it is the minimum negative anode potential given such that the photo current becomes zero.

From Einstein's relation:

maximum kinetic energy $=$$h$ν $-h$ν'

Therefore it can be written as:

$eV=h$ν$-h$ν'

#SPJ2