The stopping potential in an experiment of photon is 2eV. What is the maximum kinetic energy of photoelectrons emitted ?

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Solution

Stopping potential, $V_{o} = 2 e V$
Maximum kinetic energy, $K_{m a x} = e V_{o}$
$K_{m a x} = 1.6 \times 10^{- 19} \times 2 = 3.2 \times 10^{- 19}$ Joules

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The stopping potential in an experiment of photon is 2eV. What is the maximum kinetic energy of photoelectrons emitted ?

Stopping potential, Vo=2eVMaximum kinetic energy, Kmax=eVoKmax=1.6×10−19×2=3.2×10−19 Joules

Stopping potential, $V_{o} = 2 e V$
Maximum kinetic energy, $K_{m a x} = e V_{o}$
$K_{m a x} = 1.6 \times 10^{- 19} \times 2 = 3.2 \times 10^{- 19}$ Joules