Question

If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength $ \lambda $. When it jumps from the second to the first orbit, the corresponding wavelength of the photon will be:

• A. \( \frac{5\lambda}{27} \)
• B. \( \frac{7\lambda}{20} \)
• C. \( \frac{16\lambda}{9} \)
• D. \( \frac{20\lambda}{7} \)

Hint:

In hydrogen atom transitions, the wavelength of the emitted photon is inversely proportional to the energy difference between the orbits. Using the formula for energy levels and photon energy, we can easily determine the wavelength for different transitions.

Answer 1

The Correct Option is A

In the hydrogen atom, the energy of the electron in the \( n^{th} \) orbit is given by: \[ E_n = - \frac{13.6}{n^2} \text{ eV} \] The wavelength of the emitted photon when the electron transitions from an upper orbit \( n_2 \) to a lower orbit \( n_1 \) can be found using the relation: \[ E_{\text{photon}} = E_{n_2} - E_{n_1} \] The energy of the photon is related to the wavelength \( \lambda \) by the equation: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] where \( h \) is Planck's constant and \( c \) is the speed of light. For the transition from the 3rd orbit to the 2nd orbit, the energy difference is: \[ \Delta E_{3 \rightarrow 2} = E_3 - E_2 = - \frac{13.6}{9} - \left( - \frac{13.6}{4} \right) = \frac{13.6}{4} - \frac{13.6}{9} \] The wavelength of the emitted photon for this transition is: \[ \lambda = \frac{hc}{\Delta E_{3 \rightarrow 2}} = \frac{5}{27} \cdot \lambda \] For the transition from the 2nd orbit to the 1st orbit, the energy difference is: \[ \Delta E_{2 \rightarrow 1} = E_2 - E_1 = - \frac{13.6}{4} - \left( - \frac{13.6}{1} \right) = \frac{13.6}{1} - \frac{13.6}{4} \] The wavelength of the emitted photon for this transition is: \[ \lambda_{\text{new}} = \frac{hc}{\Delta E_{2 \rightarrow 1}} = \frac{5}{27} \cdot \lambda \] 
Thus, the wavelength of the photon emitted when the electron jumps from the second to the first orbit is \( \frac{5\lambda}{27} \).