The ratio of the deBroglie wavelengths of proton, deuteron and alpha particle accelerated through the same potential difference 100V is

Options

A. `2:2:1`
B. `1:2:2sqrt2`
C. `2:1:2`
D. `2sqrt2:2:1`

Answer

To find the ratio of the de Broglie wavelengths of a proton, deuteron, and alpha particle when they are accelerated through the same potential difference (100V), we can follow these steps:

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

### Step 2: Relate momentum to kinetic energy
The momentum \(p\) can be expressed in terms of kinetic energy (K.E.):
\[ p = \sqrt{2m \cdot \text{K.E.}} \]
When a particle is accelerated through a potential difference \(V\), the kinetic energy gained is:
\[ \text{K.E.} = qV \]
where \(q\) is the charge of the particle. Therefore, we can rewrite momentum as:
\[ p = \sqrt{2m \cdot qV} \]

### Step 3: Substitute into the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2m \cdot qV}} \]

### Step 4: Analyze the ratio of wavelengths
Since the potential difference \(V\) is the same for all three particles, we can focus on the mass \(m\) and charge \(q\) of each particle:
- For the **proton**: \(m_p = 1 \, \text{amu}\), \(q_p = 1 \, e\)
- For the **deuteron**: \(m_d = 2 \, \text{amu}\), \(q_d = 1 \, e\)
- For the **alpha particle**: \(m_{\alpha} = 4 \, \text{amu}\), \(q_{\alpha} = 2 \, e\)

### Step 5: Calculate the effective mass-charge product
We can express the ratio of the de Broglie wavelengths as:
\[ \frac{\lambda_p}{\lambda_d} = \frac{\sqrt{2m_d \cdot q_d}}{\sqrt{2m_p \cdot q_p}} = \frac{\sqrt{m_d \cdot q_d}}{\sqrt{m_p \cdot q_p}} \]
\[ \frac{\lambda_d}{\lambda_{\alpha}} = \frac{\sqrt{m_{\alpha} \cdot q_{\alpha}}}{\sqrt{m_d \cdot q_d}} \]
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\sqrt{m_{\alpha} \cdot q_{\alpha}}}{\sqrt{m_p \cdot q_p}} \]

### Step 6: Calculate the ratios
Now we can calculate the ratios:
- For the proton:
\[ \lambda_p \propto \frac{1}{\sqrt{1 \cdot 1}} = 1 \]
- For the deuteron:
\[ \lambda_d \propto \frac{1}{\sqrt{2 \cdot 1}} = \frac{1}{\sqrt{2}} \]
- For the alpha particle:
\[ \lambda_{\alpha} \propto \frac{1}{\sqrt{4 \cdot 2}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \]

### Step 7: Form the final ratio
Now we can form the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_p}{\lambda_d} : \frac{\lambda_d}{\lambda_{\alpha}} : \frac{\lambda_{\alpha}}{\lambda_p} = 1 : \frac{1}{\sqrt{2}} : \frac{1}{2\sqrt{2}} = 2\sqrt{2} : 2 : 1 \]

Thus, the final ratio of the de Broglie wavelengths of the proton, deuteron, and alpha particle is:
\[ \text{Ratio} = 2\sqrt{2} : 2 : 1 \]