Question

In the experiment on the photoelectric effect, the graph between $E_{K}$ (max) and $v$ is found to be a straight line as shown in figure. If the threshold frequency according to this graph is $x\times 10^{16}Hz$ then find $x$ .

Answer 1

$E_{K m a x}=hv-\phi_{0}$
In the graph of $E_{K m a x}$ vs $\nu$, the $Y-$ intercept represents work function $\left(\phi_{0}\right)$ and slope gives the Planck's constant $\left(\right.h\left.\right)$ .
$\therefore h=\frac{3 . 6 \times \left(10\right)^{- 15}}{\left(\right. 8 - 2 \left.\right) \times \left(10\right)^{1 / 8}}=6\times \left(10\right)^{- 34}J-s$
Now, $\phi_{0}=hv_{0}=Y-$ intercept
$\therefore hv_{0}=1.6\times 10^{- 15}$
$\therefore v_{0}=\frac{1 . 6 \times 10^{- 15}}{6 \times 10^{- 34}}=267\times 10^{16}Hz$