Question

The light emitted in the transition n = 3 to n = 2 in hydrogen is called Hα light. Find the maximum work function a metal can have so that Hα light can emit photoelectrons from it.

Answer 1

The maximum work done by the metal function is found to be 1.90 eV  so that Hα light can emit photo electrons from it.

Explanation:

If the energy of Hα light is greater than or equal to the metal’s work function, it can then emit the photo electrons.  

The amount of energy contained by Hα light (E) is shown as

$E = 13.6\times(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}} )$ eV

Here, $n_1 = 2$ and $n_2= 3$ Therefore,

$E = 13.6 \times (\frac{1}{4} - \frac{1}{9} )$

$E = \frac{13.6\times 5}{36}$

E = 1.89 eV  

E = 1.90 eV  

From the metal surface, Hα light can emit electrons for the maximum work done by the metal function and it is found to be 1.90 eV

Answer 2

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1.90eV will be the answer.