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Q.

During the transition of 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 :-

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a

$3000 Å$

b

$6000 Å$

c

$4000 Å$

d

$2000 Å$

answer is A.

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$\begin{aligned} E_{A} - E_{C} = \frac{hc}{2000 Å} \dots . . (i) \\ and E_{B} - E_{C} = \frac{hc}{6000 Å} \dots . (ii) \\ Now E_{A} - E_{B} = (E_{A} - E_{C}) - (E_{B} - E_{C}) \\ \frac{hc}{λ_{AB}} = \frac{hc}{2000} - \frac{hc}{6000} \\ \frac{1}{λ_{AB}} = \frac{1}{3000 Å} \\ λ_{AB} = 3000 Å \end{aligned}$

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