First slide

Bohr Model of the Hydrogen atom.

Question

In hydrogen atom, if the difference in the energy of the electron in n = 2 and n = 3 orbits is E, the ionization energy of hydrogen atom is :

Moderate

A

13.2 E
B

7.2 E
C

5.6 E
D

3.2 E

Solution

$g i v e n E = 13.6 (\frac{1}{2^{2}} - \frac{1}{3^{2}}) E = 13.6 (\frac{9 - 4}{36}) = \frac{13.6 x 5}{36} i o n i s a t i o n e n e r g y i s 13.6 = E \frac{36}{5} = 7.2 E$

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