### Stopping potential :

Stopping potential of kinetic energy K_{max (j)} in joules is equal to the product of electron e_{©} in coulomb and multiply the stopping potential v_{0(v)} in volts. Hence the stopping potential equation can be written as,

K_{max (j)=}e_{©}* v_{0(v)}

V_{0}= K_{max}/e

Where,

K_{max}→ kinetic energy in joules

V_{0}→ stopping potential in volts

E→ electrons in coulomb

### Example:1

Calculate the stopping potential ?In a work function of metal 3.45 ev and the light of frequency 6*10^{14} Hz is moving on the metal surface, photoemission of electrons occurs.

### Answer:

Work function of metal φ=3.45 ev

Frequency v=6.0*10^{14} HZ

As we know

K=hv – φ_{0}

Where

H=planck’s constant=6.626*10^{-34} js

K=6.626*10^{34}*6*10^{14 }/1.6 *10^{-19}-3.45

=2.485-3.450=0.965ev

Hence, the maximum kinetic energy of the emitted electron is 0.345 ev

For stopping potential as we know the equation for kinetic energy as

K=ev_{0}

V_{0}=K/e

=0.965*1.6*10^{-19} /1.6*10^{-19}=0.965v

So the stopping potential of the material is 0.965v

### Example:2

Calculate the stopping potential ? In a work function of metal 2.14 ev and the light of frequency 6*10^{14} Hz is moving on the metal surface, photoemission of electrons occurs.

### Answer:

Work function of metal φ=2.14 ev

Frequency v=6.0*10^{14} HZ

As we know

K=hv – φ_{0}

Where

H=planck’s constant=6.626*10^{-34} js

K=6.626*10^{34}*6*10^{14 }/1.6 *10^{-19}-2.14

=2.485-2.140=0.345ev

Hence, the maximum kinetic energy of the emitted electron is 0.345 ev

For stopping potential as we know the equation for kinetic energy as

K=ev_{0}

V_{0}=K/e

=0.345*1.6*10^{-19} /1.6*10^{-19}=0.345v

So the stopping potential of the material is 0.345v