If the de-Broglie wavelength is `lambda_(0)` for protons accelerated through 100 V, then the de-Broglie wavelength for alpha particles accelerated through the same voltage will be

Options

A. `lambda_(0)`
B. `(lambda_0)/(2)`
C. `(lambda_0)/(2)sqrt2`
D. none of these

Answer

To solve the problem of finding the de-Broglie wavelength for alpha particles accelerated through the same voltage as protons, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential difference
When a charged particle is accelerated through a potential difference \( V \), the kinetic energy \( K \) gained by the particle is given by:
\[ K = qV \]
where \( q \) is the charge of the particle.

### Step 2: Write down the expression for de-Broglie wavelength
The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 3: Express momentum in terms of kinetic energy
The momentum \( p \) can be expressed in terms of kinetic energy:
\[ p = \sqrt{2mK} = \sqrt{2mqV} \]
Substituting this into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]

### Step 4: Compare the de-Broglie wavelengths of protons and alpha particles
Let \( \lambda_0 \) be the de-Broglie wavelength for protons and \( \lambda' \) be the de-Broglie wavelength for alpha particles. The mass of a proton is \( m_p \) and the mass of an alpha particle (which consists of 2 protons and 2 neutrons) is \( m_{\alpha} = 4m_p \). The charge of a proton is \( q_p = e \) and the charge of an alpha particle is \( q_{\alpha} = 2e \).

Using the formula for de-Broglie wavelength:
\[ \lambda_0 = \frac{h}{\sqrt{2m_p e V}} \]
\[ \lambda' = \frac{h}{\sqrt{2(4m_p)(2e)V}} = \frac{h}{\sqrt{16m_p e V}} = \frac{h}{4\sqrt{m_p e V}} \]

### Step 5: Relate the two wavelengths
Now we can express the ratio of the two wavelengths:
\[ \frac{\lambda'}{\lambda_0} = \frac{h / (4\sqrt{m_p e V})}{h / (\sqrt{2m_p e V})} = \frac{\sqrt{2m_p e V}}{4\sqrt{m_p e V}} = \frac{1}{4} \cdot \sqrt{\frac{2}{1}} = \frac{\sqrt{2}}{4} \]
Thus,
\[ \lambda' = \frac{\sqrt{2}}{4} \lambda_0 \]

### Step 6: Final expression
Since \( \lambda_0 \) is the de-Broglie wavelength for protons, we can express \( \lambda' \) in terms of \( \lambda_0 \):
\[ \lambda' = \frac{\lambda_0}{2\sqrt{2}} \]

### Conclusion
The de-Broglie wavelength for alpha particles accelerated through the same voltage will be:
\[ \lambda' = \frac{\lambda_0}{2\sqrt{2}} \]