Question

The ionization energy of the electron in the hydrogen atom in its ground state is $13.6\, eV.$ The atoms are excited to higher energy levels to emit radiations of $6$ wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between

• A. n = 3 to n = 1 states
• B. n = 2 to n = 1 states
• C. n = 4 to n = 3 states
• D. n = 3 to n = 2 states

Answer 1

The Correct Option is C

To determine which transition corresponds to the maximum wavelength of emitted radiation from the hydrogen atom, we need to understand the relationship between energy levels and wavelengths.

The energy difference between two levels in the hydrogen atom is given by:

E = 13.6 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \, eV,

where n_1 and n_2 are the principal quantum numbers of the lower and higher energy levels, respectively.

The wavelength \lambda of the emitted radiation is inversely proportional to the energy difference:

\lambda = \frac{hc}{E},

where h is Planck’s constant and c is the speed of light.

For maximum wavelength, we need minimum energy difference. Let's calculate the energy differences for each transition:

• n = 3 to n = 1:

  E_{3 \to 1} = 13.6 \left(\frac{1}{1^2} - \frac{1}{3^2}\right) = 13.6 \left(1 - \frac{1}{9}\right) = 13.6 \times \frac{8}{9} \approx 12.09\, eV

• n = 2 to n = 1:

  E_{2 \to 1} = 13.6 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6 \left(1 - \frac{1}{4}\right) = 13.6 \times \frac{3}{4} = 10.2\, eV

• n = 4 to n = 3:

  E_{4 \to 3} = 13.6 \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = 13.6 \left(\frac{1}{9} - \frac{1}{16}\right) = 13.6 \times \frac{7}{144} \approx 0.66\, eV

• n = 3 to n = 2:

  E_{3 \to 2} = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \times \frac{5}{36} \approx 1.89\, eV

Among these transitions, the transition from n = 4 to n = 3 has the smallest energy difference of approximately 0.66\, eV, which corresponds to the maximum wavelength of emitted radiation because wavelength is inversely proportional to energy.

Therefore, the maximum wavelength of emitted radiation corresponds to the transition from n = 4 to n = 3.