The transition, so that the de–Broglie wavelength of electron becomes 3 times of its initial value in `H e ^(+)` ion will be:

Options

A. `1 to 2`
B. `2 to 5`
C. `4 to 6`
D. `2 to 6`

Answer

To solve the problem of determining the transition so that the de-Broglie wavelength of an electron in the \( He^+ \) ion becomes three times its initial value, we can follow these steps:

### Step 1: Understanding the de-Broglie Wavelength
The de-Broglie wavelength (\( \lambda \)) of an electron is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

### Step 2: Relating Momentum to Velocity
The momentum \( p \) can also be expressed in terms of mass (\( m \)) and velocity (\( v \)):
\[ p = mv \]
Thus, we can rewrite the de-Broglie wavelength as:
\[ \lambda = \frac{h}{mv} \]

### Step 3: Using Bohr's Model
According to Bohr's model, the angular momentum of an electron in a hydrogen-like atom is quantized:
\[ mvr = \frac{n h}{2\pi} \]
where \( n \) is the principal quantum number and \( r \) is the radius of the orbit.

### Step 4: Expressing Wavelength in Terms of Quantum Number
Substituting \( p = mv \) into the de-Broglie equation, we get:
\[ \lambda = \frac{h}{m \cdot \frac{n h}{2\pi r}} = \frac{2\pi r}{n} \]
Now, we need to express \( r \) in terms of \( n \) for the \( He^+ \) ion.

### Step 5: Radius of the Orbit for \( He^+ \)
For a hydrogen-like atom, the radius \( r \) of the \( n \)-th orbit is given by:
\[ r = \frac{0.529 n^2}{Z} \]
where \( Z \) is the atomic number. For \( He^+ \), \( Z = 2 \).

### Step 6: Substitute Radius into Wavelength
Substituting \( r \) into the wavelength equation:
\[ \lambda = \frac{2\pi \left(\frac{0.529 n^2}{Z}\right)}{n} = \frac{2\pi \cdot 0.529 n}{Z} \]
For \( He^+ \), this becomes:
\[ \lambda = \frac{2\pi \cdot 0.529 n}{2} = \frac{\pi \cdot 0.529 n}{1} \]

### Step 7: Setting Up the Ratio for Wavelengths
Let \( \lambda_1 \) be the initial wavelength and \( \lambda_2 \) be the final wavelength. According to the problem, we have:
\[ \lambda_2 = 3 \lambda_1 \]
Using the proportionality of \( \lambda \) to \( n \):
\[ \frac{\lambda_2}{\lambda_1} = \frac{n_2}{n_1} \]
Thus,
\[ \frac{3}{1} = \frac{n_2}{n_1} \]
This implies:
\[ n_2 = 3n_1 \]

### Step 8: Conclusion
If we assume \( n_1 = 1 \) (the ground state), then:
\[ n_2 = 3 \cdot 1 = 3 \]
Thus, the transition is from \( n = 1 \) to \( n = 3 \).

### Final Answer
The transition so that the de-Broglie wavelength of the electron becomes three times its initial value in the \( He^+ \) ion is from \( n = 1 \) to \( n = 3 \).

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