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Q. Graph of cut-off potential $\left(V_{0}\right)$ versus frequency $\left(\right.f\left.\right)$ for same intensity of light for metal plates $A$ and $B$ are as shown in the figure below. Choose the correct statement.
Question

NTA AbhyasNTA Abhyas 2022

Solution:

$hf=hf_{0}+eV_{0}$
$\therefore eV_{0}=hf-hf_{0}$
$V_{0}=\frac{h f}{e}-\frac{h f_{0}}{e}$
As $\left( \text{Intercept}\right)_{B}>\left(\text{Intercept}\right)_{A}$
$\left(\frac{h f_{0}}{e}\right)_{B}>\left(\frac{h f_{0}}{e}\right)_{A}$
$\therefore \phi_{A} < \phi_{B}$