Question

In Bohr model of hydrogen atom, the electron makes a transition from $n=5$ to $n=1$ state. As a result, a photon of wavelength $\lambda$ is emitted in the process. The wavelength of the photon emitted when an electron makes a transition from energy level $n=5$ to $n=2$ will be:

• A. $\dfrac{8}{7}\lambda$
• B. $\dfrac{24}{7}\lambda$
• C. $\dfrac{16}{7}\lambda$
• D. $\dfrac{32}{7}\lambda$

Hint:

Use 1/lambda = R(1/nf^2 - 1/ni^2) for 5 to 1 and 5 to 2, then divide the two results.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In the Bohr model, a photon is emitted when the electron falls from a higher level \( n_i \) to a lower level \( n_f \). The wavelength follows the Rydberg formula.

Step 2: Key Formula or Approach:
\[ \dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) \]
Here \( R \) is the Rydberg constant and \( Z = 1 \) for hydrogen.

Step 3: Detailed Explanation:
For the transition \( 5 \to 1 \):
\[ \dfrac{1}{\lambda} = R\left(1 - \dfrac{1}{25}\right) = \dfrac{24R}{25} \]
For the transition \( 5 \to 2 \):
\[ \dfrac{1}{\lambda_2} = R\left(\dfrac{1}{4} - \dfrac{1}{25}\right) = \dfrac{21R}{100} \]
Divide the second relation by the first to remove \( R \):
\[ \dfrac{\lambda}{\lambda_2} = \dfrac{21/100}{24/25} = \dfrac{21}{96} \]
\[ \lambda_2 = \dfrac{96}{21}\lambda = \dfrac{32}{7}\lambda \]

Step 4: Check the options.
Option (1), \( \tfrac{8}{7}\lambda \), is shorter than \( \lambda \). That is wrong because a smaller energy gap gives a longer wavelength than \( \lambda \).
Option (2), \( \tfrac{24}{7}\lambda \), equals 3.43 \( \lambda \). The correct ratio is 96/21 = 4.57, so (2) is wrong.
Option (3), \( \tfrac{16}{7}\lambda \), equals 2.29 \( \lambda \), which does not match 4.57 either.
Option (4), \( \tfrac{32}{7}\lambda \), is exactly the ratio we found.

Final Answer:
The wavelength for \( 5 \to 2 \) is \( \tfrac{32}{7}\lambda \), which is option (4). \[ \boxed{\lambda_2 = \dfrac{32}{7}\lambda} \]