Question

The stopping potential $V$ for photoelectric emission from a metal surface is plotted along $y-$axis and frequency $\gamma$ of incident light along $x-$ axis. A straight line is obtained as shown. The Planck constant is given by-

• A. Slope of the line
• B. Product of slope and mass of electron.
• C. Product of intercept along $y-$axis and mass of the electron.
• D. Product of slope of the line and charge on the electron.

Answer 1

The correct option is D Product of slope of the line and charge on the electron.

From Einstein's photoelectric equation,

$h\gamma ={\varphi }_0+eV$

$V=\left(\frac{h}{e}\right)\gamma -\frac{{\varphi }_0}{e}.....\left(1\right)$

$y=mx-c.......\left(2\right)$

Comparing (1) and (2) we get,

$\text{slope}=\frac{h}{e}$

$\therefore \text{Planck's constant}h=\text{slope}\times e$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.