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A photoelectric emitter has a threshold frequency $ϑ_{0}$ When light frequency $2 ϑ_{0}$ is incident, the speed of photo electrons is V. When light of frequency 0 10 is incident the speed of photoelectrons will be

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20V

answer is B.

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Detailed Solution

$2 {hϑ}_{0} = K . E_{1} {hϑ}_{0}$

$K . E_{1} = {hϑ}_{0} . . . . . . . . . . . . . . . . (1)$

$10 {hϑ}_{0} = K . E_{2} + {hϑ}_{0}$

$K . E_{2} = 9 {hϑ}_{0} . . . . . . . . . . . . . . . . (2)$

From eq (1) and (2)

$K . E_{2} = 9 . K . E_{1}$

$K . E_{1} = \frac{1}{2} {mV}_{1}^{2}$

$V_{1} = \sqrt{\frac{2 {KE}_{1}}{m}}$

$V_{2} = \sqrt{\frac{2 {KE}_{2}}{m}} = \sqrt{\frac{9 \times 2 {KE}_{1}}{m}} = \sqrt{\frac{2 {KE}_{1}}{3}} = 3 V$

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