The threshold frequency for a certain metal is v0 When light of frequency v = 2 v0 is incident on it, the maximum velocity of photo electrons is 4 x 106 m/s' If the frequency of incident radiation is increased to 5 v0, then the maximum velocity of photo electrons in m/s will be

A
$(4 / 5) \times 10^{6}$
B
2 x10⁶
C
8 x 10⁶
D
2 x 10⁷

Solution:

$\frac{1}{2} {mv}_{1}^{2} = 2 {hv}_{0} - {hv}_{0} = {hv}_{0} \frac{1}{2} {mv}_{2}^{2} = 5 {hv}_{0} - {hv}_{0} = 4 {hv}_{0} ∴ \frac{1}{2} m {v_{2}}^{2} = 4 [\frac{1}{2} m {v_{1}}^{2}] or v_{2} = 2 v_{1} or v_{2} = 2 \times (4 \times 10^{6}) = 8 \times 10^{6} m / s$

The threshold frequency for a certain metal is v₀ When light of frequency v = 2 v₀ is incident on it, the maximum velocity of photo electrons is 4 x 10⁶ m/s' If the frequency of incident radiation is increased to 5 v₀, then the maximum velocity of photo electrons in m/s will be

Solution:

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