For a certain metal the threshold frequency is υ0. If light of frequency 2υ0 is incident on it the electron come out with a maximum velocity of 4×106m/s. If light of frequency of 10υ0 is incident on it the maximum velocity of the photo electron will be

KE1KE2=hυ1−hυ0hυ2−hυ0⇒v1v2=υ1−υ0υ2−υ0v1v2=2ϑ0-ϑ010ϑ0-ϑ04×106v2=13v2=12×106

For a certain metal the threshold frequency is υ0. If light of frequency 2υ0 is incident on it the electron come out with a maximum velocity of 4×106m/s. If light of frequency of 10υ0 is incident on it the maximum velocity of the photo electron will be

KE1KE2=hυ1−hυ0hυ2−hυ0⇒v1v2=υ1−υ0υ2−υ0v1v2=2ϑ0-ϑ010ϑ0-ϑ04×106v2=13v2=12×106

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For a certain metal the threshold frequency is $υ_{0}$ . If light of frequency $2 υ_{0}$ is incident on it the electron come out with a maximum velocity of $4 \times 10^{6} m / s$ . If light of frequency of $10 υ_{0}$ is incident on it the maximum velocity of the photo electron will be

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$8 \times 10^{6} m / s$

$16 \times 10^{6} m / s$

$2 \times 10^{6} m / s$

$12 \times 10^{6} m / s$

answer is D.

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$\frac{K E_{1}}{K E_{2}} = \frac{h υ_{1} - h υ_{0}}{h υ_{2} - h υ_{0}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{υ_{1} - υ_{0}}{υ_{2} - υ_{0}}}$

$\frac{v_{1}}{v_{2}} = \sqrt{\frac{2 ϑ_{0} - ϑ_{0}}{10 ϑ_{0} - ϑ_{0}}}$

$\frac{4 \times 10^{6}}{v_{2}} = \frac{1}{3}$

$v_{2} = 12 \times 10^{6}$

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For a certain metal the threshold frequency is υ0. If light of frequency 2υ0 is incident on it the electron come out with a maximum velocity of 4×106m/s. If light of frequency of 10υ0 is incident on it the maximum velocity of the photo electron will be

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For a certain metal the threshold frequency is $υ_{0}$ . If light of frequency $2 υ_{0}$ is incident on it the electron come out with a maximum velocity of $4 \times 10^{6} m / s$ . If light of frequency of $10 υ_{0}$ is incident on it the maximum velocity of the photo electron will be