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Question : 12
Total: 29
In the ground state of hydrogen atom, its Bohr radius is given as 5.3 × 10 − 11 m 21.2 × 10 − 11 m
(i) the value of the principal quantum number and
(ii) the total energy of the atom in this excited state.
(i) the value of the principal quantum number and
(ii) the total energy of the atom in this excited state.
Solution:
(i) r = r 0 n 2
∴21.2 × 10 − 11 = 5.3 × 10 − 11 × n 2
∴4 = n 2
n = 2
(ii)E = −
Or, E = −
∴ E = − 3.4 eV
∴
∴
(ii)
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