First slide

Photo Electric Effect

Question

The electric field associated with a light wave is $E = E_{0} \sin [1 \cdot 57 \times 10^{7} (x - c t)]$ where x is in metre and t is in second. If this light is used to produce photo-electric emission from the surface of a metal of work-function 1'9 eV, then the stopping potential will be

Moderate

A

1.2 V
B

1.5 V
C

1.75 V
D

1.9 V

Solution

Given that
$E = E_{0} \sin [1 \cdot 57 \times 10^{7} (x - ct)] We know that, E = E_{0} \sin [\frac{2 π}{λ} (x - ct)] Comparing, we get λ = \frac{2 π}{1 \cdot 57 \times 10^{7}} m ∴ Energy of photon = \frac{hc}{λ} = \frac{(6 \cdot 64 \times 10^{- 34}) (3 \times 10^{8}) (1 \cdot 57 \times 10^{7})}{2 π \times (1 \cdot 6 \times 10^{- 19})} eV = 3 \cdot 1 eV So, maximum K . E . of photo - electron = 3 \cdot 1 - 1 \cdot 9 = 1 \cdot 2 eV This is equal to the stopping potential .$

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