Question

In Bohr model of hydrogen atom, an electron makes a transition from n = 4 state to n = 1 state and a photon of frequency $\nu$ is emitted. The frequency of photon emitted when an electron makes a transition from n = 4 state to n = 2 state in the same model is

• A. $\frac{\nu}{3}$
• B. $\frac{\nu}{4}$
• C. $\frac{\nu}{5}$
• D. $\frac{\nu}{6}$

Hint:

Using the Rydberg formula structure $\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$ is frequently the fastest and most foolproof algebraic method to compare spectral transition energies without computing absolute numerical eV values.

Answer 1

The Correct Option is C

Concept:
• When an excited electron in a hydrogen atom securely drops from a higher energy orbit ($n_i$) to a lower energy orbit ($n_f$), it strictly emits a single photon.

• The exact energy of this uniquely emitted photon is absolutely equal to the numerical energy difference between those two specific stationary states, governed by $E_{photon} = E_i - E_f$.

• The mathematical frequency of the emitted photon directly scales with this energy gap according to Planck's fundamental equation: $E_{photon} = h\nu$.

• The energy of an electron in the $n$-th Bohr orbit is effectively written as $E_n = \frac{-13.6 \text{ eV}}{n^2}$.

Step 1: Analyze the first transition ($n=4 \rightarrow n=1$)
The energy gap for an electron jumping deeply from the fourth orbit down to the ground state is:
\[ \Delta E_1 = E_4 - E_1 \]
Substitute the standard Bohr energy level formula:
\[ \Delta E_1 = \left(\frac{-13.6}{4^2}\right) - \left(\frac{-13.6}{1^2}\right) \]
\[ \Delta E_1 = 13.6 \times \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \text{ eV} \]
Calculate the fractional part carefully:
\[ \Delta E_1 = 13.6 \times \left( 1 - \frac{1}{16} \right) = 13.6 \times \left( \frac{15}{16} \right) \text{ eV} \]
The frequency $\nu$ of the emitted photon is linked directly to this energy gap:
\[ h\nu = 13.6 \times \left( \frac{15}{16} \right) \]

Step 2: Analyze the second transition ($n=4 \rightarrow n=2$)
Now, consider a different scenario where the electron drops from the fourth orbit only down to the second orbit (Balmer series).
The new energy gap is calculated similarly:
\[ \Delta E_2 = E_4 - E_2 \]
\[ \Delta E_2 = 13.6 \times \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \text{ eV} \]
Calculate the fractional part carefully:
\[ \Delta E_2 = 13.6 \times \left( \frac{1}{4} - \frac{1}{16} \right) \]
Find a common denominator to complete the strict subtraction:
\[ \Delta E_2 = 13.6 \times \left( \frac{4}{16} - \frac{1}{16} \right) = 13.6 \times \left( \frac{3}{16} \right) \text{ eV} \]
Let the unknown frequency of this newly emitted photon be $\nu'$. According to Planck's equation:
\[ h\nu' = 13.6 \times \left( \frac{3}{16} \right) \]

Step 3: Calculate the ratio of the frequencies
To rigorously find the mathematical relationship between the two frequencies, we divide the second equation strictly by the first equation:
\[ \frac{h\nu'}{h\nu} = \frac{13.6 \times \left( \frac{3}{16} \right)}{13.6 \times \left( \frac{15}{16} \right)} \]
The physical constants ($h$) and the $13.6 \text{ eV}$ scaling factor elegantly cancel out from both the numerator and the denominator:
\[ \frac{\nu'}{\nu} = \frac{\frac{3}{16}}{\frac{15}{16}} \]
The common denominator of $16$ also nicely cancels out entirely:
\[ \frac{\nu'}{\nu} = \frac{3}{15} \]
Simplify this final basic fraction:
\[ \frac{\nu'}{\nu} = \frac{1}{5} \implies \nu' = \frac{\nu}{5} \]

Step 4: Conclusion
The new frequency $\nu'$ mathematically evaluates exactly to $\frac{\nu}{5}$. This directly corresponds to option (C).