Question

Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its
(1) Second permitted energy level to the first permitted level.
(2) The highest permitted energy level to the first permitted level.

Answer 1

Hint: As a first step, one could recall the expression for energy of the photon produced as the result of electron transition. Now you could use this expression in order to find the energy of the photons produced as the result of mentioned transitions. Then, you could take the ratio of these two energies to find the answer.

Formula used:
Energy of the photon produced as the result of electron transition,
${{E}_{{{n}_{i}}\to {{n}_{f}}}}=Rhc\left( \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right)$

Complete step-by-step solution:
In the question, we are asked to find the ratio of energies of photons produced due to the electron transition of hydrogen atom between the mentioned energy levels.
(1) Energy of the photon produced due to the electron transition from second permitted energy level to the first permitted energy level.
${{E}_{2\to 1}}=Rhc\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right)=\dfrac{3}{4}Rhc$ ………………………………….. (1)
(2) Energy of the photon produced due to the electron transition from highest permitted energy level to the first permitted energy level.
${{E}_{\infty \to 1}}=Rhc\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right)=Rhc$……………………………………. (2)
Now, we are supposed to find the ratio of the above two relations, that is, equations (1) and (2).
$\dfrac{{{E}_{2\to 1}}}{{{E}_{\infty \to 1}}}=\dfrac{\dfrac{3}{4}Rhc}{Rhc}$
$\therefore \dfrac{{{E}_{2\to 1}}}{{{E}_{\infty \to 1}}}=\dfrac{3}{4}$
Therefore, we found that the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its ‘Second permitted energy level to the first permitted level’ to ‘The highest permitted energy level to the first permitted level’ to be $\dfrac{3}{4}$.


Note: You may have noticed the terms R, h and c in the solution which are the Rydberg constant, Planck’s constant and speed of light in vacuum respectively. These constants don’t play any significant role in the solution as they are getting cancelled out from the numerator and denominator while taking the ratio.