Question

The ratio of slopes of $K_{max}$ vs. $\nu$ and $V_0$ vs $\nu$ curves in the photoelectric effect gives (v =frequency, $K_{max}$ = maximum kinetic energy, $v_0$ = stopping potential):

• A. Charge of electron
• B. Planck's constant
• C. Work function
• D. The ratio of planck's constant and electronic charge

Answer 1

The correct option is A Charge of electron

$hv=h{v}_0+e{V}_0\Rightarrow e{V}_0=hv-h{v}_0\Rightarrow V_0=\frac{h}{c}v-\frac{h}{e}{v}_0\therefore {\text{slope}}_1=\frac{h}{e}$
Similarly, $hv=h{v}_0+K_{max}$
or $K_{max}=hv-h{v}_0;$
${\text{slope}}_2=h,\therefore \frac{{\text{slope}}_2}{{\text{slope}}_1}=\frac{h}{h/e}=e{}^{}{}^{\prime }{e}^{\prime }\text{is the electronic charge.}$