First slide

Bohr's atom model

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

$Energy, E = K [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] (K = constant) n_{1} = 2 and n_{2} = 3 . So , E = K [\frac{1}{2^{2}} - \frac{1}{3^{2}}] = K [\frac{5}{36}] For removing an electron n_{1} = 1 to n_{2} = \infty Energy E_{1} = K [1] = \frac{36}{5} E = 7.2 E ∴ Ionization energy = 7.2 E$

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