Using Bohr's theory, the transition, so that the electrons de-Broglie wavelength becomes 3 times of its orginial value in `He^+` ion will be

Options

A. `2to6`
B. `2to4`
C. `1to4`
D. `1to6`

Explanation

To solve the problem using Bohr's theory, we need to determine the transition in the He⁺ ion such that the de Broglie wavelength of the electron becomes three times its original value. Here’s a step-by-step solution:

### Step 1: Understanding de Broglie Wavelength
The de Broglie wavelength (λ) of an electron is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v \) is its velocity.

### Step 2: Relating Wavelength to Velocity
From the above formula, we can see that the de Broglie wavelength is inversely proportional to the velocity:
\[ \lambda \propto \frac{1}{v} \]
This means that if the wavelength increases, the velocity must decrease.

### Step 3: Velocity in Bohr's Model
In Bohr's model, the velocity \( v \) of an electron in a hydrogen-like atom is given by:
\[ v \propto \frac{Z}{n} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 4: Relating Wavelength to Quantum Numbers
Substituting the expression for velocity into the wavelength formula, we get:
\[ \lambda \propto \frac{n}{Z} \]
This shows that the wavelength is directly proportional to the principal quantum number \( n \).

### Step 5: Setting Up the Ratio
If the original wavelength is \( \lambda_1 \) and the new wavelength is \( \lambda_2 \), and we know that \( \lambda_2 = 3 \lambda_1 \), we can write:
\[ \frac{\lambda_1}{\lambda_2} = \frac{n_1}{n_2} \]
Substituting the known relationship:
\[ \frac{1}{3} = \frac{n_1}{n_2} \]
This implies:
\[ n_1 = \frac{1}{3} n_2 \]

### Step 6: Finding the Transition
Now we need to find values for \( n_1 \) and \( n_2 \) that satisfy this ratio. We can test the options provided in the question.

1. If \( n_1 = 2 \) and \( n_2 = 6 \):
\[ \frac{n_1}{n_2} = \frac{2}{6} = \frac{1}{3} \]
This satisfies our condition.

2. Testing other options:
- \( n_1 = 2 \) and \( n_2 = 4 \): \( \frac{2}{4} = \frac{1}{2} \)
- \( n_1 = 2 \) and \( n_2 = 8 \): \( \frac{2}{8} = \frac{1}{4} \)
- \( n_1 = 2 \) and \( n_2 = 12 \): \( \frac{2}{12} = \frac{1}{6} \)

Only the first option satisfies the ratio \( \frac{1}{3} \).

### Conclusion
Thus, the transition that results in the de Broglie wavelength becoming three times its original value in the He⁺ ion is from \( n = 2 \) to \( n = 6 \).

### Final Answer
The correct transition is from \( n_1 = 2 \) to \( n_2 = 6 \).

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