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For a certain metal the threshold frequency is $ν_{0}$ . If light of frequency 2 $ν_{0}$ is incident on it the electron come out with a maximum velocity of 4 × 10⁶ m/s. If light of frequency of 5 $ν_{0}$ is incident on it the maximum velocity of the photo electron will be

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8 x 10⁶ m/s

16 x 10⁶ m/s

2 x 10⁶ m/s

12 x 10⁶ m/s

answer is A.

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Detailed Solution

$\frac{1}{2}m{v^2}_{\max } = E - {W_o}\,;\,\,$

$\therefore \frac{1}{2}m{v^2}_{\max } = h(\upsilon - {\upsilon _o})$

$\therefore {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^2} = \frac{{{\upsilon _1} - {\upsilon _0}}}{{{\upsilon _2} - {\upsilon _0}}}$

$\,\left(\frac{4\times10^6}{V_2} \right )^2=\left ( \frac{2\nu_0-\nu_0}{5\nu_0-\nu_0} \right )$

$\therefore {V_2} = 8 \times {10^6}m/s\$

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