Question

During the transition of an electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 Å, and it becomes 6000 Å when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is:

• A. 3000 Å
• B. 6000 Å
• C. 4000 Å
• D. 2000 Å

Hint:

In atomic transitions, the wavelength of the emitted radiation is inversely proportional to the energy difference. Use this relationship to solve for unknown wavelengths in different transitions.

Answer 1

The Correct Option is A

Step 1: Use the Rydberg formula. The energy of a photon emitted during a transition is inversely proportional to the wavelength. 
Let $\lambda_{AC}$ be the wavelength for A to C, $\lambda_{BC}$ be the wavelength for B to C, and $\lambda_{AB}$ be the wavelength for A to B. Then, the energy relationships are: \[ \frac{1}{\lambda_{AC}} = \frac{1}{\lambda_{AB}} + \frac{1}{\lambda_{BC}} \] Step 2: Substitute the given values. We have $\lambda_{AC} = 2000$ Å and $\lambda_{BC} = 6000$ Å. We want to find $\lambda_{AB}$. \[ \frac{1}{2000} = \frac{1}{\lambda_{AB}} + \frac{1}{6000} \] Step 3: Solve for $\lambda_{AB}$. \[ \frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000} = \frac{3}{6000} - \frac{1}{6000} = \frac{2}{6000} = \frac{1}{3000} \] \[ \lambda_{AB} = 3000 { Å} \] Therefore, the wavelength of the radiation emitted during the transition of electrons from state A to state B is 3000 Å. The correct answer is (1).