Question

The electron in a hydrogen atom makes a transition ${n_1} \to {n_2}$, where ${n_1}$ and ${n_2}$ are the principal quantum numbers of the two energy states. Assume Bohr's model to be valid. The time period of the electron in the initial state is eight times that in the final state. What are the possible values of ${n_1}$ and ${n_2}$?

Answer 1

Before calculating the ${n_1}$ and ${n_2}$ values, you should know the relation between time period and orbit of the electron. We all know the time period is equal to the revolution taking per frequency. So we can use this relation to obtain the values of ${n_1}$ and ${n_2}$.

Formula used:
$T = \dfrac{1}{f} \propto {n^3}$
Where, $T$ - is the time period,
$f$ -frequency
$n$ -number of orbitals

Complete step by step solution:
The period for the ${n_1}$ th orbitals electron is written as,
${T_1} = \dfrac{1}{{{f_1}}} \propto {n_1}^3$
The period for the ${n_2}$ th orbitals electron is written as,
${T_2} = \dfrac{1}{{{f_2}}} \propto {n_2}^3$
Hence, divide the two time period, we get
$\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{n_1^3}}{{n_2^3}}$
The time period for ${n_1}$ orbital is $8$ times greater than the ${n_2}$ orbital. So we can write the relation as,
${T_1} = 8{T_2}$,
Substitute the relation into the equation means, we can get,
$8 = {(\dfrac{{{n_1}}}{{{n_2}}})^3}$
$\Rightarrow {n_1} = 2{n_2}$
The possible values of ${n_1}$ and ${n_2}$ are, ${n_1} = 2;{n_2} = 1$, ${n_1} = 4;{n_2} = 2$, ${n_1} = 6;{n_2} = 3$ and so on.

Note: From Bohr's model of the hydrogen atom, easily calculate the time period taken to complete the one rotation of the electron. The time period is calculated from the simple formula, the time period is equal to the total distance covered divided by the velocity of the electron. From this formula also we can easily calculate the time period of Bohr's orbit. The time it takes to complete one rotation to the electron is $1.6 \times {10^{ - 16}}$ seconds.