Q28 / 120 · VITEEE 2015 · Chemistry

The ratio of slopes of $K_{\max }$ vs $V$ and $V_{0}$ vs $v$ 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
  • Correct answer
    D
    charge of electron

Solution

$\begin{array}{l} \mathrm{hv}=\mathrm{hv}_{0}+\mathrm{ev}_{0} \\ \mathrm{v}_{0}=\frac{\mathrm{h}}{\mathrm{e}} \mathrm{v}-\frac{\mathrm{h}}{\mathrm{e}} \mathrm{v}_{0} \end{array}$ On comparing this equation with the straight line equation, i.e $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ The slope of $\mathrm{v}_{0}$ vs $\mathrm{v}$ is ( $v_{0}$ is stopping potential) $(\text { slope })_{1}=\frac{\mathrm{h}}{\mathrm{e}}$ Likewise, $\mathrm{hv}=\mathrm{hv}_{0}+\mathrm{K}_{\max }$ $\text { or } \quad \mathrm{K}_{\max }=\mathrm{h} \mathrm{v}-\mathrm{h} \mathrm{v}_{0}$ Thus, slope of $K_{\max }$ vs $v$ is $(\text { slope })_{2}=\mathrm{h} \therefore \frac{(\text { slope })_{2}}{(\text { slope })_{1}}=\frac{\mathrm{h}}{\mathrm{h} / \mathrm{e}}=\mathrm{e}$

Hence, the correct answer is option: 'D'.

Relevant Questions