In a photoelectric experiment for $4000 \circ A$
incident radiation, the potential difference to stop the ejection is 2 V. If the incident light is changed to $3000 \circ A$
, then the potential required to stop the ejection of electrons will be

2 V

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Less than 2 V

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Zero

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Greater than 2 V

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Solution

The correct option is D
Greater than 2 V

According to Einstein's photoelectric equation
$E = W_{0} + K_{m a x} \Rightarrow V_{0} = \frac{h c}{e} [\frac{1}{λ} - \frac{1}{λ_{0}}]$
Hence if $λ$ decreases $V_{0}$ increases

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In a photoelectric experiment for 4000∘A incident radiation, the potential difference to stop the ejection is 2 V. If the incident light is changed to 3000∘A , then the potential required to stop the ejection of electrons will be

The correct option is D Greater than 2 V According to Einstein's photoelectric equation E=W0+Kmax⇒V0=hce[1λ−1λ0] Hence if λ decreases V0 increases

In a photoelectric experiment for $4000 \circ A$
incident radiation, the potential difference to stop the ejection is 2 V. If the incident light is changed to $3000 \circ A$
, then the potential required to stop the ejection of electrons will be

The correct option is D
Greater than 2 V

According to Einstein's photoelectric equation
$E = W_{0} + K_{m a x} \Rightarrow V_{0} = \frac{h c}{e} [\frac{1}{λ} - \frac{1}{λ_{0}}]$
Hence if $λ$ decreases $V_{0}$ increases