What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n = 4 to n = 2 of He+ spectrum?

Explanation

To solve the problem of finding which transition in the hydrogen spectrum has the same wavelength as the Balmer transition from n = 4 to n = 2 in the He+ spectrum, we can follow these steps:

### Step 1: Understand the Formula
We start with the formula for the wavelength of transitions in hydrogen-like atoms:
\[ \frac{1}{\lambda} = Z^2 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( Z \) is the atomic number, \( R_H \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

### Step 2: Calculate for He+
For the He+ ion (which has \( Z = 2 \)):
- The transition is from \( n_2 = 4 \) to \( n_1 = 2 \).
- Plugging these values into the formula gives:
\[ \frac{1}{\lambda_{He^+}} = 2^2 R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ = 4 R_H \left( \frac{1}{4} - \frac{1}{16} \right) \]
\[ = 4 R_H \left( \frac{4 - 1}{16} \right) = 4 R_H \left( \frac{3}{16} \right) = \frac{3}{4} R_H \]

### Step 3: Calculate for Hydrogen
For hydrogen (which has \( Z = 1 \)):
- We need to find \( n_1 \) and \( n_2 \) such that:
\[ \frac{1}{\lambda_{H}} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
We equate this to the previous result:
\[ R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3}{4} R_H \]
Dividing both sides by \( R_H \):
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4} \]

### Step 4: Solve for \( n_1 \) and \( n_2 \)
Assuming \( n_1 < n_2 \), we can try \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]
This satisfies the equation.

### Conclusion
The transition in the hydrogen spectrum that has the same wavelength as the Balmer transition from \( n = 4 \) to \( n = 2 \) in the He+ spectrum is from \( n = 2 \) to \( n = 1 \).