An `alpha` particle and a proton have their masses in the ratio 4:1 and charges in the ratio 2: 1. Find ratio of their de-Broglie wavelengths when both move with equal velocities.
In Q-20, find ratio of de-Broglie wavelengths, when both have equal kinetic energy.

Options

A. `2:1`
B. `1:1`
C. `4:1`
D. `1:4`

Answer

To find the ratio of the de-Broglie wavelengths of an alpha particle and a proton when both have equal kinetic energy, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (λ) of a particle 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 \( K \):
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the particle and \( K \) is its kinetic energy.

### Step 3: Substitute momentum into the de-Broglie wavelength formula
Substituting the expression for momentum into the de-Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]

### Step 4: Set up the ratio of de-Broglie wavelengths
Let \( \lambda_{\alpha} \) be the wavelength of the alpha particle and \( \lambda_{p} \) be the wavelength of the proton. Thus, we have:
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha}K}} \quad \text{and} \quad \lambda_{p} = \frac{h}{\sqrt{2m_{p}K}} \]

### Step 5: Find the ratio of the wavelengths
To find the ratio \( \frac{\lambda_{\alpha}}{\lambda_{p}} \):
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{\sqrt{2m_{p}K}}{\sqrt{2m_{\alpha}K}} = \frac{\sqrt{m_{p}}}{\sqrt{m_{\alpha}}} \]

### Step 6: Substitute the mass ratio
We know that the mass ratio of the alpha particle to the proton is given as:
\[ \frac{m_{\alpha}}{m_{p}} = 4 \implies m_{\alpha} = 4m_{p} \]
Thus,
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{\sqrt{m_{p}}}{\sqrt{4m_{p}}} = \frac{\sqrt{m_{p}}}{2\sqrt{m_{p}}} = \frac{1}{2} \]

### Step 7: Final ratio of de-Broglie wavelengths
Therefore, the ratio of the de-Broglie wavelengths of the alpha particle to the proton is:
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{1}{2} \implies \lambda_{\alpha} : \lambda_{p} = 1 : 2 \]