Question 152: Atoms & Nuclei

Question

During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is $2000 \mathop A\limits^o$ and it becomes $6000 \mathop A\limits^o$ 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

Accepted Answer

$$ \frac{1}{\lambda_{AC}} = \frac{E_A - E_C}{hc}, \quad \frac{1}{\lambda_{BC}} = \frac{E_B - E_C}{hc} $$
Since the electron transitions from state A to state C in two steps (A to B and then B to C), the energy difference for the transition from A to B is given by
$$ E_A - E_B = (E_A - E_C) - (E_B - E_C) $$
Expressing these energy differences in terms of wavelengths,
$$ \frac{hc}{\lambda_{AB}} = \frac{hc}{\lambda_{AC}} - \frac{hc}{\lambda_{BC}} $$
Cancelling out $hc$, we obtain
$$ \frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}} $$
Substitute the given values:
$$ \frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000} $$
Finding a common denominator:
$$ \frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000} $$
Thus, the wavelength for the A to B transition is
$$ \lambda_{AB} = 3000 \, \mathring{A}. $$
This matches Option D.

Answer

$4000 \mathop A\limits^o$

Answer

$6000 \mathop A\limits^o$

Answer

$2000 \mathop A\limits^o$

Answer

$3000 \mathop A\limits^o$

Atoms & Nuclei

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