Question

The ratio of minimum to maximum wavelength of radiation emitted by transition of an electron to ground state of Bohr's hydrogen atom is:
A. $\dfrac{3}{4} \\ $
B. $\dfrac{1}{4} \\ $
C. $\dfrac{1}{8} \\ $
D. $\dfrac{3}{8}$

Answer 1

Hint:As the transition is from $n = \infty $ to $n = 1$. So it is considered in the Lyman series as the final state is $n=1$. The energy of the radiation produced increases with the difference between the degrees of transition, resulting in a decrease in wavelength. To solve the question, we need to use the energy formula, which is expressed in terms of the wavelength, Planck's constant, and the speed of light in a vacuum.

Formula used: The expression of the energy of the radiation corresponds to the transition between two energy levels is given by:
$E = 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)eV$
And, the energy which is necessary for the transition of an electron is given by:
$E = \dfrac{{hc}}{\lambda }$
Here, $h$ is the Planck constant and $c$ is the speed of light.

Complete step by step solution:
As we know that the Energy of the radiation corresponds to the transition between two energy levels ${n_1}$ and ${n_2}$ is given by:
$E = 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)eV \\ $
Now, the transition of the electron must occur in a way that releases the least amount of energy in order for the maximal wavelength to be emitted. The ground state of Bohr's hydrogen atom is ${n_1} = 1$, the just-next-energy state, which is ${n_2} = 2$, would be the nearest energy state for the minimum energy, so we can substitute the values in the above formula, then we have:
${E_{\min }} = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)eV \\$
$\Rightarrow {E_{\min }} = 13.6\left( {1 - \dfrac{1}{4}} \right)eV \\$
$\Rightarrow {E_{\min }} = 13.6 \times \dfrac{3}{4}eV \\$
Similarly, when the atom is ionised then the ground state of Bohr's hydrogen atom is ${n_1} = 1$, the just-next-energy state, which is ${n_2} = \infty $, would be the nearest energy state for the maximum energy, so we can substitute the values in the above formula, then we have:
${E_{\max }} = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)eV \\$
$\Rightarrow {E_{\max }} = 13.6\left( {1 - \dfrac{1}{\infty }} \right)eV \\$
$\Rightarrow {E_{\max }} = 13.6eV \\$
As per the question, the proportion of an electron's transition to the ground state of the hydrogen atom proposed by Bohr's by its minimum to maximum wavelength of radiation. The energy released on each of the transitions mentioned in the question must now be compared to this value, so we obtain:
$\dfrac{{{E_{\min }}}}{{{E_{\max }}}} = \dfrac{{13.6 \times \dfrac{3}{4}eV}}{{13.6eV}}\,\,\,\,\,...(i)$

To identify the wavelength, we can disregard the negative sign, which suggests that energy was released during the transition. Then, we can keep in mind that the energy and wavelength are related by:
$E = \dfrac{{hc}}{\lambda }\,\,\,\,\,....(ii)$
Now, substitute the equation $(ii)$ in the equation $(i)$, we obtain:
$\dfrac{{{{\left( {\dfrac{{hc}}{\lambda }} \right)}_{\min }}}}{{{{\left( {\dfrac{{hc}}{\lambda }} \right)}_{\max }}}} = \dfrac{{13.6 \times \dfrac{3}{4}eV}}{{13.6eV}} \\$
$\therefore \dfrac{{{\lambda _{\min }}}}{{{\lambda _{\max }}}} = \dfrac{3}{4} $
Therefore, the radiation's emission's wavelength ratio is $\dfrac{3}{4}$.

Thus, the correct option is A.

Note: It should be noted that an EMR transition's energy is inversely proportional to wavelength and directly proportional to frequency. Therefore, we may conclude that the energy of the EMR will increase with greater frequency and decrease with a higher wavelength of the EMR.