Answer :

The maximum kinetic energy of the emitted electron is equal to work done on the electron to stop it to reach to the anode. Therefore


K.Emax = 1/2 mv2max = eV0


Magnitude of charge of the electron ‘e’ = 1.6 × 10-19 C


V0 = Stopping Potential


K.Emax = maximum kinetic energy of emitted electron


m = mass of the electron = 9.1 × 10-31 kg


vmax = maximum velocity of electron


K.Emax = eV0 = 1.6 × 10-19 × 1.5 = 1.5 Ev


= 2.4 × 10-19 J


maximum KE of electron is 2.4 × 10-19 J


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