Question

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

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

Answer 1

Answer: (A)

According to Einstein's photoelectric equation
$hv=(K.E{)}_{\max }+{\varphi }_0$
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
$eV=hv-{\varphi }_0$
$V=\frac{h}{e}v-\frac{{\varphi }_0}{e}$
Hence, the graph between $V$ and $v$ is a straight line and slope of this graph is given by
Slope $=\frac{h}{e}\dots (i)$
From the graph in the question, slope $=\frac{ab}{bc}\dots (ii)$
From (i) and (ii), we get $h=e\frac{ab}{bc}$