When a light of frequency $v_{1}$ is incident on a metal surface, the photoelectrons emitted have twice the kinetic energy as did the photoelectrons emitted when the same metal has irradiated with light of frequency $v_{2}$ . What will be the value of threshold frequency?

v_{0} = v_{1} - v_{2}

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v_{0} = v_{1} - 2 v_{2}

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v_{0} = 2 v_{2} - v_{1}

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v_{0} = v_{1} + v_{2}

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Solution

The correct option is C $v_{0} = 2 v_{2} - v_{1}$
$h v_{1} = h v_{0} + K E_{1}$ .....(1)
$h v_{2} = h v_{0} + K E_{2}$ ....(2)

But $K E_{1} = 2 K E_{2}$ .....(3)
Substitute equation (3) in equation (1)
$h v_{1} = h v_{0} + 2 K E_{2}$ ....(4)

From equation (2)
$h v_{2} - h v_{0} = K E_{2}$ ....(5)

Substitute equation (5) in equation (4)
$h v_{1} = h v_{0} + 2 (h v_{2} - h v_{0})$
$h v_{1} = h v_{0} + 2 h v_{2} - 2 h v_{0}$

$h v_{1} = 2 h v_{2} - h v_{0}$
$h v_{0} = 2 h v_{2} - h v_{1}$
$v_{0} = 2 v_{2} - v_{1}$

Hence, the value of threshold frequency will be $v_{0} = 2 v_{2} - v_{1}$ .

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