The longest wavelength of the Lyman series for hydrogen atom is the same as the wavelength of a certain line in the spectrum of `He^+` when the electron makes a trensiton from `n rarr 2`. The value of n is

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To solve the problem, we need to find the value of \( n \) for the transition in the spectrum of \( He^+ \) that corresponds to the longest wavelength of the Lyman series for the hydrogen atom. Let's break down the solution step by step.

### Step 1: Understand the Lyman Series
The Lyman series corresponds to transitions in the hydrogen atom where the electron falls to the \( n = 1 \) energy level from higher energy levels (\( n \geq 2 \)). The longest wavelength in the Lyman series occurs when the transition is from \( n = 2 \) to \( n = 1 \).

### Step 2: Calculate the Wavelength for Hydrogen
Using the Rydberg formula for hydrogen:
\[ \frac{1}{\lambda_H} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the transition \( n_2 = 2 \) to \( n_1 = 1 \):
\[ \frac{1}{\lambda_H} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus,
\[ \lambda_H = \frac{4}{3R} \]

### Step 3: Wavelength for Helium Ion \( He^+ \)
For the \( He^+ \) ion, we use the same Rydberg formula, but with \( Z = 2 \):
\[ \frac{1}{\lambda_{He}} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the transition from \( n \) to \( n_2 = 2 \):
\[ \frac{1}{\lambda_{He}} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{n^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{n^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda_{He}} = R \left( 1 - \frac{4}{n^2} \right) \]

### Step 4: Set the Wavelengths Equal
Since the problem states that the wavelengths are equal:
\[ \lambda_H = \lambda_{He} \]
Taking the inverse:
\[ \frac{1}{\lambda_H} = \frac{1}{\lambda_{He}} \]
Substituting the expressions we derived:
\[ R \left( \frac{3}{4} \right) = R \left( 1 - \frac{4}{n^2} \right) \]
Cancelling \( R \) from both sides:
\[ \frac{3}{4} = 1 - \frac{4}{n^2} \]

### Step 5: Solve for \( n^2 \)
Rearranging the equation:
\[ \frac{4}{n^2} = 1 - \frac{3}{4} = \frac{1}{4} \]
Cross-multiplying gives:
\[ 4 = \frac{4}{n^2} \implies n^2 = 16 \]
Thus,
\[ n = 4 \]

### Final Answer
The value of \( n \) is \( 4 \).

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