First slide

Photo Electric Effect

Question

A photon of energy 8 eV is incident on metal surface of threshold frequency 1.6 x $10^{15}$ Hz. The maximum kinetic energy of the photoelectrons emitted (in eV) (Take h = 6 x $10^{- 34}$ Js).

Moderate

A

1.6
B

6
C

2
D

1.2

Solution

$\begin{array}{l} K . E . = hv - {hv}_{o} = 8 eV \\ = 8 eV - (\frac{8 \times 10^{- 34} \times 1 .6 \times 10^{15}}{1 .6 \times 10^{- 19}} eV) \\ = 8 eV - 6 eV = 2 eV \end{array}$

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