Q93 / 120 · JEE Advanced 2006 · Physics

The graph between $1 / \lambda$ and stopping potential $(V)$ of three metals having work function $\phi_1, \phi_2$ and $\phi_3$ in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? [Here $\lambda$ is the wavelength of the incident ray]. Figure

  • Correct answer
    A
    Ratio of work functions $\phi_1: \phi_2: \phi_3=1: 2: 4$.
  • B
    Ratio of work functions $\phi_1: \phi_2: \phi_3=4: 2: 1$.
  • Correct answer
    C
    $\tan \theta$ is directly proportional to $h c / e$, where $h$ is Planck's constant and $c$ is the speed of light.
  • D
    The violet colour light can eject photoelectrons from metals 2 and 3.

Solution

From the relation, $e V=\frac{h c}{\lambda}-\phi$ or $V=\left(\frac{h c}{e}\right)\left(\frac{1}{\lambda}\right)-\frac{\phi}{e}$ This is equation of straight line. Slope is $\tan \theta=\frac{h c}{e}$ $$ \begin{aligned} \phi_1: \phi_2: \phi_3 & =\frac{h c}{\lambda_{01}}: \frac{h c}{\lambda_{02}}: \frac{h c}{\lambda_{03}}=\frac{1}{\lambda_{01}}: \frac{1}{\lambda_{02}}: \frac{1}{\lambda_{03}}=1: 2: 4 \\ \frac{1}{\lambda_{01}} & =0.001 \mathrm{~nm}^{-1} \text { or } \lambda_{01}=10000 Å \\ \frac{1}{\lambda_{02}} & =0.002 \mathrm{~nm}^{-1} \text { or } \lambda_{02}=5000 Å \\ \frac{1}{\lambda_{03}} & =0.004 \mathrm{~nm}^{-1} \text { or } \lambda_{03}=25000 Å \end{aligned} $$ Violet colour has wavelength $4000 Å$. So, violet colour can eject photoelectrons from metal-1 and metal-2.

Hence, the correct answer is options: 'A', 'C'.

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