Question

Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.

• A. 3 : 4
• B. 4 : 3
• C. 1 : 4
• D. 4 : 1

Answer 1

The Correct Option is A

To find the ratio of energies of photons produced due to the electron transitions in a hydrogen atom, we need to use the formula for energy in the Bohr model of the hydrogen atom:

E_n = -\frac{13.6 \, \text{eV}}{n^2}

where n is the principal quantum number, and 13.6 eV is the ionization energy of the hydrogen atom.

Let's solve for both scenarios given in the question:

1. For the transition from the second permitted energy level (n=2) to the first level (n=1):

   The energy of the photon is given by the difference in energy levels:

   \Delta E_{21} = E_1 - E_2 = \left(-\frac{13.6}{1^2}\right) - \left(-\frac{13.6}{2^2}\right)

   \Delta E_{21} = -13.6 + \frac{13.6}{4}

   \Delta E_{21} = -13.6 + 3.4 = -10.2 \, \text{eV}

2. For the transition from the highest permitted energy level (n=\infty) to the first level:

   The energy of the photon for a transition from infinity to the first level is:

   \Delta E_{\infty 1} = E_1 - E_\infty = \left(-\frac{13.6}{1^2}\right) - 0 = -13.6 \, \text{eV}

The ratio of the energies of the photons produced due to these transitions is:

\frac{\Delta E_{21}}{\Delta E_{\infty 1}} = \frac{-10.2 \, \text{eV}}{-13.6 \, \text{eV}}

\Rightarrow \frac{10.2}{13.6} = \frac{3}{4}

Thus, the required ratio is 3 : 4.