MHT CET · Physics · Dual Nature of Matter
In case of photoelectric effect, the graph of measured stopping potential \(\left(\mathrm{V}_0\right)\) against frequency ' \(v\) ' of incident light is a straight line. The slope of this line multiplied by the charge of electron (e) gives
- A the work function of the metal.
- B the Planck's constant.
- C the maximum kinetic energy of the ejected electrons.
- D the threshold frequency for photoejection from the metal.
Answer & Solution
Correct Answer
(B) the Planck's constant.
Step-by-step Solution
Detailed explanation
We know that,
\(\begin{aligned}
& \mathrm{V}_0=\frac{\mathrm{h} \nu}{\mathrm{l}}-\frac{\mathrm{h} \mathrm{v}_0}{l} \\
& \mathrm{y}=\mathrm{mx}+\mathrm{c}
\end{aligned}\)
The slope of this line is \(\frac{\mathrm{h}}{l}\)
According to the question,
\(\frac{\mathrm{h}}{l} \times \mathrm{e}\)
\(\therefore \quad\) The slope of line is ' \(h\) ' (Planck's constant).
\(\begin{aligned}
& \mathrm{V}_0=\frac{\mathrm{h} \nu}{\mathrm{l}}-\frac{\mathrm{h} \mathrm{v}_0}{l} \\
& \mathrm{y}=\mathrm{mx}+\mathrm{c}
\end{aligned}\)
The slope of this line is \(\frac{\mathrm{h}}{l}\)
According to the question,
\(\frac{\mathrm{h}}{l} \times \mathrm{e}\)
\(\therefore \quad\) The slope of line is ' \(h\) ' (Planck's constant).
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