Question

The ratio of slopes of $K_{max}$ vs $V$ and $V_0$ vs $v$ curves in the photoelectric effects gives: [$v=$ frequency, $K_{max}=$ maximum kinetic energy, $v_0=$ stopping potential]

• A. the ratio of Planck's constant of electronic charge
• B. work function
• C. Planck's constant
• D. charge of electron

Answer 1

The correct option is D charge of electron

$hv=h{v}_0+e{v}_0$

$v_0=\frac{h}{e}v-\frac{h}{e}{v}_0$

On comparing this equation with the straight line equation, i.e $y=mx+c$

The slope of $v_0$ vs $v$ is ($v_0$ is stopping potential)
${\left(slope\right)}_1=\frac{h}{e}$

or $K_{max}=hv-h{v}_0$

Thus, slope of $K_{max}$ vs $v$ is

${\left(slope\right)}_2=h$

$\therefore$ $\frac{{\left(slope\right)}_2}{{\left(slope\right)}_1}=\frac{h}{h/e}=e$

Hence, the correct option is $D$