What will be the wavelength of radiation released in nano-metre when an electron drops from the 5th Bohr orbit to 2nd Bohr orbit in hydrogen atom?

Options

A. 217nm
B. 434nm
C. 651nm
D. 868nm

Explanation

To find the wavelength of radiation released when an electron drops from the 5th Bohr orbit to the 2nd Bohr orbit in a hydrogen atom, we can use the Rydberg formula for hydrogen. Here’s a step-by-step solution:

### Step 1: Understand the Rydberg Formula
The Rydberg formula for the wavelength of emitted radiation during an electron transition in a hydrogen atom is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \(\lambda\) is the wavelength,
- \(R\) is the Rydberg constant (\(R = 109678 \, \text{cm}^{-1}\)),
- \(n_1\) is the lower energy level,
- \(n_2\) is the higher energy level.

### Step 2: Identify the Energy Levels
In this case:
- The electron drops from the 5th orbit (\(n_2 = 5\)) to the 2nd orbit (\(n_1 = 2\)).

### Step 3: Substitute Values into the Rydberg Formula
Substituting the values into the formula:
\[ \frac{1}{\lambda} = 109678 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \]
Calculating \( \frac{1}{2^2} \) and \( \frac{1}{5^2} \):
\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{5^2} = \frac{1}{25} = 0.04 \]
Now, substituting these values:
\[ \frac{1}{\lambda} = 109678 \left( 0.25 - 0.04 \right) = 109678 \times 0.21 \]

### Step 4: Calculate \(\frac{1}{\lambda}\)
Calculating:
\[ \frac{1}{\lambda} = 109678 \times 0.21 = 23032.38 \, \text{cm}^{-1} \]

### Step 5: Find \(\lambda\)
To find \(\lambda\), take the reciprocal:
\[ \lambda = \frac{1}{23032.38} \approx 0.0000434 \, \text{cm} \]
To convert this to nanometers (1 cm = \(10^7\) nm):
\[ \lambda \approx 0.0000434 \, \text{cm} \times 10^7 \, \text{nm/cm} = 434 \, \text{nm} \]

### Final Answer
The wavelength of radiation released when an electron drops from the 5th Bohr orbit to the 2nd Bohr orbit in a hydrogen atom is approximately **434 nm**.

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