Question

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

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

Answer 1

Answer: (D)

n = 2 to n = 1

Answer 2

The frequency ($\nu$) of radiation emitted during an electronic transition in a hydrogen atom is given by the formula: $h\nu = E_{n_i} - E_{n_f}$ where $h$ is Planck's constant, $E_{n_i}$ is the energy of the initial state, and $E_{n_f}$ is the energy of the final state. The energy of an electron in the $n$-th orbit is given by $E_n = -\frac{13.6}{n^2}$ eV. For emission, the electron transitions from a higher energy level ($n_i$) to a lower energy level ($n_f$), so $n_i > n_f$. The energy difference is: $\Delta E = E_{n_i} - E_{n_f} = \left(-\frac{13.6}{n_i^2}\right) - \left(-\frac{13.6}{n_f^2}\right) = 13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{ eV}$ The frequency is then: $\nu = \frac{\Delta E}{h} = \frac{13.6}{h} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ To find the maximum frequency, we need to maximize the term $\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$. This term is maximized when $n_f$ is as small as possible and $n_i$ is as large as possible. The smallest possible value for $n_f$ is 1.

Let's evaluate the term $\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ for each given transition:

1. $n = 3$ to $n = 2$: $n_i = 3, n_f = 2$. Term = $\frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$.
2. $n = 5$ to $n = 4$: $n_i = 5, n_f = 4$. Term = $\frac{1}{4^2} - \frac{1}{5^2} = \frac{1}{16} - \frac{1}{25} = \frac{25-16}{400} = \frac{9}{400}$.
3. $n = 4$ to $n = 3$: $n_i = 4, n_f = 3$. Term = $\frac{1}{3^2} - \frac{1}{4^2} = \frac{1}{9} - \frac{1}{16} = \frac{16-9}{144} = \frac{7}{144}$.
4. $n = 2$ to $n = 1$: $n_i = 2, n_f = 1$. Term = $\frac{1}{1^2} - \frac{1}{2^2} = \frac{1}{1} - \frac{1}{4} = \frac{4-1}{4} = \frac{3}{4}$.

Comparing the values: $\frac{5}{36} \approx 0.1389$ $\frac{9}{400} = 0.0225$ $\frac{7}{144} \approx 0.0486$ $\frac{3}{4} = 0.75$

The largest value is $\frac{3}{4}$, which corresponds to the transition from $n = 2$ to $n = 1$. Therefore, this transition will have the maximum frequency.