Question

The ratio of slopes of $K_{\max } \operatorname{Vs} V$ and $V_{o} V S \vee$ curves in the photoelectric effects gives $\left(v=\right.$ 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

Answer: (D)

$h v=h v_{0}+e v_{0}$
or, $e v_{0}=h v-h v_{0}$
or, $v_{0}=\frac{h}{e} v-\frac{h}{e} v_{0}$
On comparing the above equation with the straight line equation, i.e $y=m x+c$ The slope of $v_{0}$ vs $v$ is
$\text { (slope })_{1}=\frac{h}{e}$
Similarly,
Thus, Slope of $\ce{K_{max} \, vs \, {v}}$ is
$\ce{ (slope)_2 = h \, \, \therefore \frac{(slope)_2}{(slope)_1} = \frac{h}{h/e} = e }$