First slide

Photo Electric Effect

Question

For intensity I of a light of wavelength 5000 $\overset{0}{A}$ the photoelectron saturation current is 0.40 μA and stopping potential is 1.36 V, the work function of the metal is:

Moderate

A

2.47 eV
B

1.36 eV
C

1.10 eV
D

0.43 eV

Solution

By using $E = W_{0} + K_{max}$
$E = \frac{12375}{5000} = 2 .475 eV and$
$K_{max} = {eV}_{0} = 1 .36 eV$
So
$2 .475 = W_{0} + 1 .36 \Rightarrow W_{0} = 1 .1 eV .$

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