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A photosensitive metallic surface has work function $h\nu_0$. If photons of energy $2h\nu_0$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^{6}$ m/s. When the photon energy is increased to $5h\nu_0$, then maximum velocity of photo electrons will be
A
$2 \times 10^{7}$ m/s
B
$2 \times 10^{6}$ m/s
C
$8 \times 10^{5}$ m/s
D
$8 \times 10^{6}$ m/s
Detailed Solution
Einstein's equation: photon energy = work function + $\dfrac{1}{2}mv_{max}^2$
First case: $2h\nu_0 = h\nu_0 + \dfrac{1}{2}m(4 \times 10^{6})^2 \Rightarrow \dfrac{1}{2}m(4 \times 10^{6})^2 = h\nu_0$
Second case: $5h\nu_0 = h\nu_0 + \dfrac{1}{2}mv^2 \Rightarrow \dfrac{1}{2}mv^2 = 4h\nu_0$
So $\dfrac{1}{2}mv^2 = 4 \times \dfrac{1}{2}m(4 \times 10^{6})^2$
$v^2 = 4 \times (4 \times 10^{6})^2$
$v = 2 \times 4 \times 10^{6} = 8 \times 10^{6}$ m/s
First case: $2h\nu_0 = h\nu_0 + \dfrac{1}{2}m(4 \times 10^{6})^2 \Rightarrow \dfrac{1}{2}m(4 \times 10^{6})^2 = h\nu_0$
Second case: $5h\nu_0 = h\nu_0 + \dfrac{1}{2}mv^2 \Rightarrow \dfrac{1}{2}mv^2 = 4h\nu_0$
So $\dfrac{1}{2}mv^2 = 4 \times \dfrac{1}{2}m(4 \times 10^{6})^2$
$v^2 = 4 \times (4 \times 10^{6})^2$
$v = 2 \times 4 \times 10^{6} = 8 \times 10^{6}$ m/s