Question

In a photoelectric experiment, the graph of frequency $\upsilon$ of incident light (in Hz) and stopping potential V (in V) is as shown in the figure. From figure, the value of the Plancks constant is (e is the elementary charge)

• A. $e\frac{ab}{bc}$
• B. $e\frac{cb}{ab}$
• C. $e\frac{ac}{bc}$
• D. $e\frac{ac}{ab}$

Answer 1

The correct option is A $e\frac{ab}{bc}$

According to Einstein's photoelectronic equation $h\mu$ = $(K.E{)}_{max}+{\varphi }_0$
where ${\varphi }_0$ is the work function , $\upsilon$ is the incident frequency and h is the Planck's constant
Also, $(K.E.{)}_{max}=eV$
where e is the elementary charge, V is the stopping potential
$eV=h\upsilon -{\varphi }_0\Rightarrow V=\frac{h}{e}\upsilon -\frac{{\varphi }_0}{e}$
Hence, the graph between V and $\upsilon$ is a straight line and slope of this graph is given by
Slope = $\frac{h}{e}$
From the graph in the question
Slope = $\frac{ab}{bc}$
From (i) and (ii) , we get , $h=e\frac{ab}{bc}$