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

To solve this problem, we apply the concept of energy level transitions in the Bohr model of the atom. When an electron transitions between energy levels, it emits or absorbs radiation. The wavelength of the radiation is related to the energy difference between the initial and final states by the equation:

\(E = \dfrac{hc}{\lambda}\) 

where:

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

Given:

• 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: \(\lambda_{BC} = 6000 \text{ Å}\)

We need to find the wavelength of the radiation emitted during the transition from state A to state B, denoted as \(\lambda_{AB}\).

To find this, we use the relationship between these wavelengths implied by Bohr's model:

• The energy difference between states is additive, as these are quantized transitions: \(E_{AB} = E_{AC} - E_{BC}\).

Now, substituting the known wavelengths into the energy relations gives:

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

Thus, 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 gives us:

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

Finding a common denominator:

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

Thus, \(\lambda_{AB} = 3000 \text{ Å}\), which means the correct option is 3000 Å.

Answer 2

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).