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

12mv12=2hv0−hv0=hv0 12mv22=5hv0−hv0=4hv0 ∴ 12mv22=412mv12 or v2=2v1 or v2=2×4×106=8×106m/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

12mv12=2hv0−hv0=hv0 12mv22=5hv0−hv0=4hv0 ∴ 12mv22=412mv12 or v2=2v1 or v2=2×4×106=8×106m/s

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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

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$(4 / 5) \times 10^{6}$

8 x 10⁶

2 x 10⁷

2 x10⁶

answer is C.

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Detailed Solution

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