Question
The light emitted in the transition n = 3 to n = 2 in hydrogen is called $\text{H}_\alpha$ light. Find the maximum work function a metal can have so that $\text{H}_\alpha$ light can emit photoelectrons from it.
✓
Answer
$n_1 = 2, n_2 = 3$ Energy possessed by $\text{H}_\alpha$ light$=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Big)$
$=13.6\times\Big(\frac{1}{4}-\frac{1}{9}\Big)$
$=1.89\text{ev}$
For $\text{H}_\alpha$ light to be able to emit photoelectrons from a metal the work function must be greater than or equal to 1.89ev.
$=13.6\times\Big(\frac{1}{4}-\frac{1}{9}\Big)$
$=1.89\text{ev}$
For $\text{H}_\alpha$ light to be able to emit photoelectrons from a metal the work function must be greater than or equal to 1.89ev.
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