Question

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state.
The ratio of the wavelengths \(\lambda_1 : \lambda_2\) emitted in the two cases is:

• A. $\frac{7}{5}$
• B. $\frac{27}{20}$
• C. $\frac{27}{5}$
• D. $\frac{20}{7}$

Answer 1

The Correct Option is D

 To determine the ratio of the wavelengths emitted when an electron in a hydrogen atom transitions from the third excited state to the second excited state and then from the second excited to the first excited state, we need to use the formula for the wavelength of the photon emitted during an electron transition in a hydrogen atom:

The change in energy when an electron transitions from a higher energy level \( n_i \) to a lower energy level \( n_f \) is given by:

\(E = -13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\text{ eV}\)

Where: 
- \( n_i \) = initial energy level (quantum number), - \( n_f \) = final energy level (quantum number), - 13.6 eV is the ionization energy of hydrogen.

The wavelength of the emitted photon can be found using:

\(\lambda = \frac{hc}{E}\)

where \( h \) is Planck's constant and \( c \) is the speed of light. However, since we are looking for a ratio, these constants will cancel out in our calculations.

Transitions:

1. First Transition: From the third excited state (n=4) to the second excited state (n=3).
   • Initial level \( n_i = 4 \)
   • Final level \( n_f = 3 \)
   • The energy difference is: \(\Delta E_1 = -13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = -13.6 \left( \frac{1}{9} - \frac{1}{16} \right) \text{ eV}\)
   • Simplifying gives: \(\Delta E_1 = -13.6 \left( \frac{16 - 9}{144} \right) = -13.6 \left(\frac{7}{144}\right) \text{ eV}\)
2. Second Transition: From the second excited state (n=3) to the first excited state (n=2).
   • Initial level \( n_i = 3 \)
   • Final level \( n_f = 2 \)
   • The energy difference is: \(\Delta E_2 = -13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = -13.6 \left( \frac{1}{4} - \frac{1}{9} \right) \text{ eV}\)
   • Simplifying gives: \(\Delta E_2 = -13.6 \left( \frac{9 - 4}{36} \right) = -13.6 \left(\frac{5}{36}\right) \text{ eV}\)

Calculating the Wavelength Ratio:

The ratio of the wavelengths is inversely proportional to the ratio of the energies, because: \(\lambda_1 = \frac{hc}{\Delta E_1}\) and \(\lambda_2 = \frac{hc}{\Delta E_2}\)

Thus, the ratio becomes: \(\frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{\frac{5}{36}}{\frac{7}{144}}\)

By solving this, we get:

• Calculate: \(= \frac{5}{36} \times \frac{144}{7} = \frac{5 \times 144}{36 \times 7} = \frac{720}{252} = \frac{20}{7}\)

 

Thus, the ratio of the wavelengths \( \lambda_1 : \lambda_2 \) is \(\frac{20}{7}\).