Question

According to Bohr atom model, in which of the following transitions will the frequency be maximum ?

• A. n = 2 to n = 1
• B. n = 3 to n = 2
• C. n = 4 to n = 3
• D. n = 5 to n = 4

Hint:

The first transition of any series (e.g., $n=2$ to $n=1$ in Lyman) always has much higher energy than any transition between higher adjacent levels.

Answer 1

The Correct Option is A

To determine which transition has the maximum frequency in the Bohr atom model, we need to use the concept of energy difference between two levels and how it relates to frequency.

1. According to the Bohr model, the energy of an electron in the n^{th} orbit is given by: E_n = - \frac{13.6 \, \text{eV}}{n^2}.
2. The frequency of the radiation emitted (or absorbed) is related to the energy difference between two levels by the relation: \Delta E = h \nu, where \nu is the frequency and h is Planck's constant.
3. The energy difference between two levels n_i and n_f is given by: \Delta E = E_{n_f} - E_{n_i} = -13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{eV}.
4. To find the maximum frequency, we need the maximum energy difference:
   • Transition from n = 2 to n = 1: \Delta E = -13.6 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 10.2 \text{eV}.
   • Transition from n = 3 to n = 2: \Delta E = -13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 1.89 \text{eV}.
   • Transition from n = 4 to n = 3: \Delta E = -13.6 \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = 0.66 \text{eV}.
   • Transition from n = 5 to n = 4: \Delta E = -13.6 \left(\frac{1}{4^2} - \frac{1}{5^2}\right) = 0.31 \text{eV}.
5. From these calculations, the energy difference is highest for the transition from n = 2 to n = 1 (10.2 eV), leading to the highest frequency.

Therefore, the transition which will result in the maximum frequency is from n = 2 to n = 1.