Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# A Hydrogen Atom in a State Having a Binding Energy of 0.85 Ev Makes Transition to a State with Excitation Energy 10.2 E.V (A) Identify the Quantum Numbers N of the Upper and the - Physics

Sum

A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 e.V (a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.

#### Solution

(a) The binding energy of hydrogen is given by

E = 13.6/(n^2)eV

For binding energy of 0.85 eV,

n_2^2 = 13.6/0.85

n_2 = 4

For binding energy of 10.2 eV,

n_2^2 = 13.6/10.2 = 16

n2 = 1.15

⇒ n1 = 2

The quantum number of the upper and the lower energy state are 4 and 2, respectively.

(b) Wavelength of the emitted radiation (λ) is given by

1/lamda = R (1/n_1^2 - 1/n_2^2)

Here,

R = Rydberg constant

n1 and n2 are quantum numbers.

therefore 1/lamda = 1.097 xx 10^7 (1/4 -  1/16)

rArr lamda = (16)/(1.097xx3xx10^7)

= 4 .8617 × 10-7

= 487 nm

Concept: The Line Spectra of the Hydrogen Atom
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 21 Bohr’s Model and Physics of Atom
Q 12 | Page 384