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. $e \frac{a b}{c b}$
• B. $e \frac{c b}{a b}$
• C. $e \frac{a c}{b c}$
• D. $e \frac{a c}{a b}$

Answer 1

Answer: (A)

According to Einstein's photoelectric equation
$h v=( K . E )_{\max }+\phi_{0}$
where $\phi_{0}$ 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 v-\phi_{0} $
$V=\frac{h}{e} v-\frac{\phi_{0}}{e}$
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 $=\frac{a b}{b c}\,\,\,\,\,\,\dots(ii)$
From (i) and (ii), we get $h=e \frac{ab}{bc}$