The hydrogen atom in its ground state is excited by means of monochromatic radiation. Its resulting spectrum has six different lines. These radiations are incident on a metal plate. It is observed that only two of them are responsible for photoelectric effect. If the ratio of maximum kinetic energy of photoelectrons in the two cases is 5 then find the work function of the metal.

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$1.93 e V$

$19.3 e V$

$0.5 e V$

$11.93 e V$

answer is D.

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Detailed Solution

Given $\frac{K_{1}}{K_{2}} = 5$ ,

$\Rightarrow \frac{Δ E_{1} - W}{Δ E_{2} - W} = 5 \dots (i)$

Since there are six different spectral lines, electrons are transitioning from n=2, n=3, and n=4 states to ground state (n=1).

Here, $Δ E_{1} = E_{4} - E_{1} = - 0.85 - (- 13.6) = 12.75 e V$ ,

and $Δ E_{2} = E_{3} - E_{1} = - 1.51 - (- 13.6) = 12.09 e V$ .

Substituting in equation (1) and solving we get

$W = \frac{5 ∆ E_{2} - ∆ E_{1}}{4} = \frac{5 \times 12.09 - 12.75}{4} = 11.93 e V$

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