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

Answer: 267

$E_{Kmax}=hv-{\varphi }_0$
In the graph of $E_{Kmax}$ vs $\nu$, the $Y-$ intercept represents work function $\left({\varphi }_0\right)$ and slope gives the Planck's constant $(h)$ .
$\therefore h=\frac{3.6\times {\left(10\right)}^{-15}}{(8-2)\times {\left(10\right)}^{1/8}}=6\times {\left(10\right)}^{-34}J-s$
Now, ${\varphi }_0=h{v}_0=Y-$ intercept
$\therefore h{v}_0=1.6\times 10^{-15}$
$\therefore v_0=\frac{1.6\times 10^{-15}}{6\times 10^{-34}}=267\times 10^{16}Hz$