Question

In a certain electronic transition in a hydrogen atom from an initial state to a final state, the difference of orbit radius is $8$ times the first Bohr radius. Which transition will satisfy the given condition?
(A) $7 \to 1$
(B) $6 \to 1$
(C) $5 \to 1$
(D) $3 \to 1$

Answer 1

Hint : Given that an electronic transition in a hydrogen atom occurs with the difference of orbit radius, which is $8$ times the first Bohr radius. The transition can be known by comparing the radius of two orbits. The first Bohr orbit means the first energy level.

Complete Step By Step Answer:
The hydrogen is the element with atomic number $1$ . An atom consists of energy levels like electronic, vibrational, rotational and translational energy levels.
The radius of the orbit will be given as:
${r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mZ{e^2}}}$
r is the radius of orbit
n is the number of transition state from which transition occurs
h is Planck’s constant
m is the mass of orbit
Z is atomic number
e is charge on proton
Given the electronic transition in hydrogen atoms from an initial state to a final state, the difference of orbit radius is $8$ times the first Bohr radius.
The radius of two orbits is ${r_{{n_2}}} - {r_{{n_1}}}$
This difference is equal to $8$
Substitute the above formula in the difference of two radius
${r_{{n_2}}} - {r_{{n_1}}} = \dfrac{{{n_2}^2{h^2}}}{{4{\pi ^2}mZ{e^2}}} - {r_n} = \dfrac{{{n_1}^2{h^2}}}{{4{\pi ^2}mZ{e^2}}} = 8$
To satisfy the above relation, the values of ${n_2}$ and ${n_1}$ are $3$ and $1$ .
The transition is $3 \to 1$
Option D is the correct one.

Note :
The terms in radius of orbit are constant except the number of orbits from which the transition occurs. As the transition involves the same atom, the atomic number and mass of orbit are also constant. The radius of orbit can be easily calculated from the value of energy level.