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The threshold frequency for a certain metal is $\nu_0$ . When a certain radiation of frequency 2 $\nu_0$ is incident on this metal surface the maximum velocity of the photoelectrons emitted is 2 × 10⁶ ms^–1. If a radiation of frequency 3 $\nu_0$ is incident on the same metal surface the maximum velocity of the photoelectrons emitted (in ms^–1 ) is

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$2\sqrt 2 \times \,{10^6}\$

$4\sqrt 3 \times \,{10^6}\$

2 x 10⁶

$4\sqrt 2 \times \,{10^6}\$

answer is B.

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Detailed Solution

\,\frac{1}{2}m{V^2}_{\max } = h(\nu - {\nu _o})\

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

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