Question

When a high frequency electromagnetic radiation is incident on a metallic surface, electrons are emitted from the surface. Energy of emitted photoelectrons depends only on the frequency of incident electromagnetic radiation and the number of emitted electrons depends only on the intensity of incident light.

Einstein's photoelectric equation $[K_{max}=h\nu -\varphi ]$ correctly explains the PE, where $\nu =$ frequency of incident light and $\varphi =$ work function.

The slope of the graph shown in fig. [here h is the Planck's constant and e is the charge of an electron] is

• A. $\frac{h}{e}$
• B. $eh$
• C. $h$
• D. $\frac{e}{h}$

Answer 1

The correct option is A $\frac{h}{e}$

We have : Using photoelectric Effect

$E_{gnc}=\varphi +K{E}_{max}....\left(1\right)$

where $E_{gnc}=$ Incident energy

$=hv$

$p=$ work function

$K{E}_{max}=Max.k.Energy$ of emitted photoelectrons

where $K{E}_{max}=e{V}_o$, where

$V_o$ is the stopping potential

Hence $e{V}_o=E_{gnc}-\varphi$

$\Rightarrow e{V}_o=\frac{hc}{\lambda }-\varphi$

$V_o=\frac{hc}{\lambda e}-\frac{\varphi }{e}$

$\Rightarrow V_o=\frac{hv}{e}-\frac{\varphi }{e}...\left(2\right)$

$\Rightarrow V_o\frac{h}{e}\left(\frac{c}{\lambda }\right)-\frac{\varphi }{e}$

Hence Curve between stopping potential and source frequency represent $\frac{h}{e}$ as the slope from the equation $\left(2\right)$