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 Planck’s 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

Answer: (A)

According to Einsteins photoelectric equation
$h\upsilon =\left(K.E\right)_{max}+\phi_{0}$
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
$eV=h\upsilon-\phi_{0}
\Rightarrow V=\frac{h}{e} \upsilon-\frac{\phi_{0}}{e}$
Hence, the graph between $V$ and $\upsilon$ is a straight line and slope of this graph is given by
Slope $=\frac{h}{e}$ $\dots (i)$
From the graph in the question
Slope $= \frac{ab}{bc}$ $\dots (ii)$
From (i) and (ii), we get, $h=e \frac{ab}{bc}$