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

The solution utilizes the Bohr model's principle of energy level transitions, where electron shifts between energy levels result in the emission or absorption of radiation. The relationship between the radiation's wavelength and the energy difference between the initial and final states is given by:

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

In this equation:

• \(E\) represents the energy difference between the two states.
• \(h\) is Planck's constant.
• \(c\) is the speed of light.
• \(\lambda\) is the wavelength of the emitted or absorbed radiation.

The provided data includes:

• Wavelength of radiation emitted from state A to state C: \(\lambda_{AC} = 2000 \text{ Å}\)
• Wavelength of radiation emitted from state B to state C: \(\lambda_{BC} = 6000 \text{ Å}\)

The objective is to determine the wavelength of the radiation emitted during the transition from state A to state B, denoted as \(\lambda_{AB}\).

Bohr's model implies the following relationship for these wavelengths:

• Energy differences are additive for quantized transitions: \(E_{AB} = E_{AC} - E_{BC}\).

Substituting the given wavelengths into the energy relations yields:

• \(E_{AC} = \dfrac{hc}{2000 \text{ Å}}\)
• \(E_{BC} = \dfrac{hc}{6000 \text{ Å}}\)

Therefore, the energy difference for the transition from A to B is:

\(E_{AB} = \dfrac{hc}{2000 \text{ Å}} - \dfrac{hc}{6000 \text{ Å}}\)

Combining these terms results in:

\(E_{AB} = hc \left(\dfrac{1}{2000} - \dfrac{1}{6000}\right)\)

Using a common denominator:

\(E_{AB} = hc \left(\dfrac{3-1}{6000}\right) = hc \left(\dfrac{2}{6000}\right) = \dfrac{hc}{3000 \text{ Å}}\)

Consequently, the wavelength \(\lambda_{AB} = 3000 \text{ Å}\). The correct option is 3000 Å.