Question

In a photo electric experiment 4 electrons with varying kinetic energy come out.

$E_1=1.6\times {10}^{-19}J{E}_2=2.1\times {10}^{-19}J{E}_3=3.2\times {10}^{-19}J{E}_4=1.2\times {10}^{-19}J$

The stopping potential , given this data will be Volt

Answer 1

Stopping potential is the potential required to stop the highest energy electrons from just hitting the collector.

The highest energy electron in the given setup is $E3=3.2\times {10}^{-19}J$

and energy is conserved , that we know for sure.

Initially the electron has kinetic energy = $3.2\times {10}^{-19}J$

Finally it stops, so final K. E = 0

K.E. lost = $3.2\times {10}^{-19}J$ …(i)

P.E. gained = $q\Delta V$

= $-1.6\times {10}^{-19}(0–V_0)$

= $(1.6\times {10}^{-19})V_0$ …(ii)

$3.2\times {10}^{-19}=1.6\times {10}^{-19}{V}_0$

$V_0=2volt.$