The energy of the electron, the hydrogen atom, is known to be expressible in the form
$E_{n} = \frac{- 13.6 eV}{n^{2}} (n = 1, 2, 3, . . . .)$
Use this expression to show that the
(i) electron in the hydrogen atom can not have an energy of – 2 V.
(ii) spacing between the lines (consecutive energy levels) within the given set of the observed hydrogen atom spectrum decreases as n increases.

Solution

Energy for the hydrogen atom is given by,

$E_{n} = - \frac{13.6 eV}{n^{2}}$

Putting $n = 1, 2, 3 . . . . . . . . . . . . n, we get$

$E_{1} = \frac{- 13.6}{12} = - 13.6 eV E_{2} = \frac{- 13.6}{2^{2}} = - \frac{13.6}{4} = 3.4 eV E_{3} = \frac{- 13.6}{3^{2}} = - \frac{13.6}{9} = - 1.51 eV$

$E_{4} = \frac{- 13.6}{4^{2}} = \frac{- 13.6}{16} = - . 85 V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E_{n} = \frac{- 13.6}{\infty^{2}} = 0 eV$

(i) Hence, from the above expressions it can be observed that the electron in the hydrogen atom cannot have an energy of – 2V.

(ii) As n increases, energies of the excited states come closer and closer together. Therefore, as n increases, E_n becomes less negative until at n = ∞, we get E_n = 0.

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