A photon of energy $4 e V$ is incident on a metal surface whose work function is $2 e V$ . The minimum reverse potential to be applied for stopping the emission of electrons is-

2 V

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4 V

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6 V

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8 V

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Solution

The correct option is A $2 V$
We know that,
$\frac{1}{2} {m V}_{m a x}^{2} = h v - ϕ_{2} = 4 e V - 2 e V = 2 e V$ where $V =$ frequency & $h =$ speed
$∴$ $e V_{0} = 2 e V$
$∴ V_{0} = 2 V$

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A photon of energy 4 eV is incident on a metal surface whose work function is 2 eV. The minimum reverse potential to be applied for stopping the emission of electrons is-

The correct option is A 2VWe know that,12mV2max=hv−ϕ2=4eV−2eV=2eV where V= frequency & h= speed∴ eV0=2eV∴V0=2V

A photon of energy $4 e V$ is incident on a metal surface whose work function is $2 e V$ . The minimum reverse potential to be applied for stopping the emission of electrons is-

The correct option is A $2 V$
We know that,
$\frac{1}{2} {m V}_{m a x}^{2} = h v - ϕ_{2} = 4 e V - 2 e V = 2 e V$ where $V =$ frequency & $h =$ speed
$∴$ $e V_{0} = 2 e V$
$∴ V_{0} = 2 V$