First slide

Photo Electric Effect

Question

Energy required to remove an electron from an aluminum surface is 4.2 eV, lf light of wavelength 2000 A falls on the surface, the velocity of fastest electron erected from the surface is

Moderate

A

2.8 x10¹⁸ m/s
B

2.5 x10¹³m/s
C

6.7 x10¹⁸m/s
D

none of these

Solution

$\frac{1}{2} {mv}^{2} = \frac{hc}{λ} - W_{0} \frac{1}{2} {mv}^{2} = \frac{(6 \cdot 6 \times 10^{- 3^{4}}) (3 \times 10^{\circ})}{(2000 \times 10^{- 10}) \times (1 \cdot 6 \times 10^{- 19})} - 4 \cdot 2 eV = 6 \cdot 2 eV - 4 \cdot 2 eV = 2 eV = 2 \times (1 \cdot 6 \times 10^{- 19}) = 3 \cdot 2 \times 10^{- 19} J Now v = \sqrt{[\frac{2 \times (3 \cdot 2 \times 10^{- 19})}{(9 \cdot 1 \times 10^{- 31})}]} = 0 \cdot 703 \times 10^{6} m / s$

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