- From the given figure, we have
Threshold frequency of A, v0A = 5 × 1014Hz and
Threshold frequency of B, v0B = 10 × 1014Hz
Now,Work function, $\phi=\text{hv}_0$
$\Rightarrow\ \phi_0\propto\text{v}_0$ [$\because$ h (Planck's constant) is constant]
Now, $\frac{\phi_{0\text{A}}}{\phi_{0\text{B}}}=\frac{5\times10^{14}}{10\times10^{14}}$
$\Rightarrow\ \frac{\phi_{0\text{A}}}{\phi_{0\text{B}}}=\frac{1}{2}$
$\therefore\ {\phi_{0\text{A}}}<{\phi_{0\text{B}}}$
Hence, the work function of B is higher than A.
- For metal A,
The slope is given by $\frac{\text{h}}{\text{e}}=\frac{2}{(10-5)10^{14}}$
$\Rightarrow\ \text{h}=\frac{2\text{e}}{5\times10^{14}}$
$=\frac{2\times1.6\times10^{-19}}{5\times10^{14}}\ \ \big[\because\ \text{e}=1.6\times10^{-19}\big]$
$=6.4\times10^{-34}\text{Js}$
Similarly, for metal B,
The slopw is given by slope $\frac{\text{h}}{\text{e}}=\frac{2.5}{(15-10)10^{14}}$
$\Rightarrow\ \text{h}=\frac{2.5\times\text{e}}{5\times10^{14}}$
$=\frac{2.5\times1.6\times10^{-19}}{5\times10^{14}}\ \ \big[\because\ 1.6\times10^{-19}\big]$
$=8\times10^{-34}\text{js}$
Since, the experimental value of h for metals A and B is different.
Hence, the given experiment is not consistent with theory.
