Single Choice

In a photocell circuit the stopping potential, $$v_0$$ , is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency v of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $$ e= 1.6 \times 10^{-19} C $$ )

A$$ 6.4 \times 10^{-34} Js, 2.0$$ eV
B$$ 6.0 \times 10^{-34} Js, 2.0$$ eV
Correct Answer
C$$ 6.4 \times 10^{-34} Js, 3.2 $$ eV
D$$ 6.0 \times 10^{-34} Js, 3.2 $$ eV

Solution

Observing the graph and relating it to the Einstein photoelectric equation, we have the graph between stopping potential and frequency as shown in the figure.

The slope of the graph shown is $$\frac{4-(-2)}{(1.6-0)\times 10^{15}}$$
Thus, $$\frac{h}{e}=\frac{6}{1.6\times 10^{15}} \Rightarrow h = 6\times 10^{-34} Js$$

Work function is given by the negative value of Y-intercept.

Y-intercept in the figure is $$-2V \Rightarrow w=2eV$$

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