During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 Aᵒ and it becomes 6000 Aᵒ 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

Solution

Physics sir · 3 steps

Add up energies, not wavelengths: A to C is the same as A to B followed by B to C.

1. Energy adds up. A to C has the bigger energy (shorter wavelength) than B to C, so B lies between A and C. The jump A to C can be split into A to B, then B to C, so E_(AC) = E_(AB) + E_(BC).
2. Switch to wavelength. Each photon has E = hc/λ. Cancel hc from every term and the energy sum becomes a sum of reciprocals of wavelength.

   1/λ_(AB) = 1/λ_(AC) - 1/λ_(BC) = 1/2000 - 1/6000 = 2/6000 = 1/3000

3. So. λ_(AB) = 3000 Å, which is option (d).

The idea to keep. When transitions chain together, the photon energies add, so 1/λ adds. Never add or subtract the wavelengths themselves.

Answer from the NTA final answer key. Solution by PrepWiser.