For certain metal the threshold frequency is ν0 , if light of frequency 2ν0 is incident on it, the electrons come out with a maximum velocity of 4×106m/s . If light of frequency of 5ν0 is incident on it, the maximum velocity of the photoelectron will be n×106 m/s where n is __

E=w0+12mv2⇒12mv2=E−w0v1v22=E1−w0E2−w0 v1v22=2hν0−hν05hν0−hν0=14v1v2=14=12⇒v2=2v1 v2=2×4×106=8×106 m/s

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For certain metal the threshold frequency is $ν_{0}$ , if light of frequency $2 ν_{0}$ is incident on it, the electrons come out with a maximum velocity of $4 \times 10^{6} m / s$ . If light of frequency of $5 ν_{0}$ is incident on it, the maximum velocity of the photoelectron will be $n \times 10^{6} m / s$ where $n$ is __

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Detailed Solution

$\begin{array}{l} E = w_{0} + \frac{1}{2} m v^{2} \Rightarrow \frac{1}{2} m v^{2} = E - w_{0} \\ {(\frac{v_{1}}{v_{2}})}^{2} = \frac{E_{1} - w_{0}}{E_{2} - w_{0}} \\ {(\frac{v_{1}}{v_{2}})}^{2} = \frac{2 h ν_{0} - h ν_{0}}{5 h ν_{0} - h ν_{0}} = \frac{1}{4} \\ \frac{v_{1}}{v_{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \Rightarrow v_{2} = 2 v_{1} \\ v_{2} = 2 \times 4 \times 10^{6} = 8 \times 10^{6} m / s \end{array}$