Q. 33.6( 20 Votes )

# The photoelectric

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.E_{max} = 1/2 mv^{2}_{max} = eV_{0}

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

V_{0} = Stopping Potential

K.E_{max} = maximum kinetic energy of emitted electron

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

v_{max} = maximum velocity of electron

K.E_{max} = eV_{0} = 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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