Threshold frequency for a certain metal is v0. When light of frequency 2v0 is incident on it, the maximum velocity of photoelectrons is 4 x 108 cm s-1. If frequency of incident radiation is increased to 5v0, then the maximum velocity of photoelectrons, in cm s-1, will be

Einstein's equation for photoelectric effect ishv−hv0=12mvmax2 When v=2v0,vmax=4×108cms−12hv0−hv0=(1/2)m4×1082hvo=12m4×1082 .......(i) v=5v0,vmax=v′h5v0−hv0=12mv′2 ......(ii)Dividing Eq. (ii) by Eq. (i), we getv′=8×108cms−1

Threshold frequency for a certain metal is v0. When light of frequency 2v0 is incident on it, the maximum velocity of photoelectrons is 4 x 108 cm s-1. If frequency of incident radiation is increased to 5v0, then the maximum velocity of photoelectrons, in cm s-1, will be

Einstein's equation for photoelectric effect ishv−hv0=12mvmax2 When v=2v0,vmax=4×108cms−12hv0−hv0=(1/2)m4×1082hvo=12m4×1082 .......(i) v=5v0,vmax=v′h5v0−hv0=12mv′2 ......(ii)Dividing Eq. (ii) by Eq. (i), we getv′=8×108cms−1

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Threshold frequency for a certain metal is v₀. When light of frequency 2v₀ is incident on it, the maximum velocity of photoelectrons is 4 x 10⁸ cm s^-1. If frequency of incident radiation is increased to 5v₀, then the maximum velocity of photoelectrons, in cm s^-1, will be

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$20 \times 10^{8}$

$8 \times 10^{8}$

$(4 / 5) \times 10^{8}$

$2 \times 10^{8}$

answer is C.

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Detailed Solution

Einstein's equation for photoelectric effect is

$\begin{array}{l} h v - h v_{0} = \frac{1}{2} m v_{max}^{2} \\ When v = 2 v_{0}, v_{max} = 4 \times 10^{8} {cms}^{- 1} \\ 2 h v_{0} - h v_{0} = (1 / 2) m {(4 \times 10^{8})}^{2} \\ h v_{o} = \frac{1}{2} m {(4 \times 10^{8})}^{2} . . . . . . . (i) \end{array}$