Question
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Answer
The frequency of metal P is lower than Q , due to which value of threshold wavelength will be higher. Since,
$
\lambda_0=\frac{c}{v}
$Therefore, the threshold frequency of plate P will be greater. The value of frequency for plate $Q$ is high and as work function $= h \times$ threshold frequency
So work function of metal Q will be high.
$
\lambda_0=\frac{c}{v}
$Therefore, the threshold frequency of plate P will be greater. The value of frequency for plate $Q$ is high and as work function $= h \times$ threshold frequency
So work function of metal Q will be high.
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