Let the energy of an emitted photoelectron be E and the wave-length of incident light be $λ$ . What will be the change in E if $λ$ is doubled?

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E/2

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E/4

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Solution

The correct option is B E/2
We have
$h ν = W_{0} + E$ where E is the energy of emitted photoelectron
$\Rightarrow \frac{h c}{λ} = W_{0} + E$
As $h c$ and $W_{0}$ are constant.
$E \propto \frac{1}{λ}$
Therefore, as $λ$ is doubled, E will become half.

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Let the energy of an emitted photoelectron be E and the wave-length of incident light be λ. What will be the change in E if λ is doubled?

The correct option is B E/2We havehν=W0+E where E is the energy of emitted photoelectron⇒hcλ=W0+EAs hc and W0 are constant.E∝1λTherefore, as λ is doubled, E will become half.

Let the energy of an emitted photoelectron be E and the wave-length of incident light be $λ$ . What will be the change in E if $λ$ is doubled?

The correct option is B E/2
We have
$h ν = W_{0} + E$ where E is the energy of emitted photoelectron
$\Rightarrow \frac{h c}{λ} = W_{0} + E$
As $h c$ and $W_{0}$ are constant.
$E \propto \frac{1}{λ}$
Therefore, as $λ$ is doubled, E will become half.