Question

The ratio of slopes of $K_{max}vs\nu$ and $V_{0}vs\nu$ curves in the photoelectric effect gives:
$(\nu$ = 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

We know that,
$h\nu =h{\nu }_0+e{V}_0$
$e{V}_0=h\nu -h{\nu }_0$
$\text{or},$
$V_0=\frac{h}{e}\nu -\frac{h}{e}{\nu }_0$
${\text{Slope}}_1=\frac{h}{e}$
Similarly, $h\nu =h{\nu }_0+K_{max}$
$\text{or},$
$K_{max}=h\nu -h{\nu }_0;$
${\text{Slope}}_2=h,$

$\frac{{\text{slope}}_2}{{\text{slope}}_1}=\frac{h}{h/e}=e$