A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of `lambda`.An alpha particle having certain kinetic energy has the same De-Brogle wavelength `lambda`.The ratio of kinetic energy of proton and that of alpha particle is:

Options

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

Answer

To solve the problem, we need to find the ratio of the kinetic energy of a proton to that of an alpha particle, given that both have the same de Broglie wavelength.

### Step-by-step Solution:

1. **Understanding de Broglie Wavelength**:
The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) can be expressed as \( p = mv \), where \( m \) is the mass and \( v \) is the velocity of the particle.

2. **Setting up the equation for the proton**:
For the proton, we have:
\[ \lambda = \frac{h}{m_p v_p} \]
where \( m_p \) is the mass of the proton and \( v_p \) is its velocity.

3. **Setting up the equation for the alpha particle**:
For the alpha particle, we have:
\[ \lambda = \frac{h}{m_\alpha v_\alpha} \]
where \( m_\alpha \) is the mass of the alpha particle and \( v_\alpha \) is its velocity.

4. **Equating the two wavelengths**:
Since both particles have the same de Broglie wavelength \( \lambda \):
\[ \frac{h}{m_p v_p} = \frac{h}{m_\alpha v_\alpha} \]
We can cancel \( h \) from both sides:
\[ \frac{1}{m_p v_p} = \frac{1}{m_\alpha v_\alpha} \]
Rearranging gives:
\[ m_p v_p = m_\alpha v_\alpha \]

5. **Relating the masses of the particles**:
The mass of an alpha particle is approximately 4 times that of a proton:
\[ m_\alpha = 4 m_p \]
Substituting this into the equation gives:
\[ m_p v_p = 4 m_p v_\alpha \]
Dividing both sides by \( m_p \) (assuming \( m_p \neq 0 \)):
\[ v_p = 4 v_\alpha \]

6. **Given the velocity of the proton**:
The problem states that the proton is moving with one-tenth the velocity of light:
\[ v_p = \frac{c}{10} \]
Therefore, substituting this into the equation gives:
\[ \frac{c}{10} = 4 v_\alpha \]
Solving for \( v_\alpha \):
\[ v_\alpha = \frac{c}{40} \]

7. **Calculating the kinetic energies**:
The kinetic energy \( KE \) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
For the proton:
\[ KE_p = \frac{1}{2} m_p \left(\frac{c}{10}\right)^2 = \frac{1}{2} m_p \frac{c^2}{100} \]
For the alpha particle:
\[ KE_\alpha = \frac{1}{2} m_\alpha \left(\frac{c}{40}\right)^2 = \frac{1}{2} (4 m_p) \frac{c^2}{1600} = \frac{2 m_p c^2}{1600} = \frac{m_p c^2}{800} \]

8. **Finding the ratio of kinetic energies**:
Now, we can find the ratio of the kinetic energies:
\[ \frac{KE_p}{KE_\alpha} = \frac{\frac{1}{2} m_p \frac{c^2}{100}}{\frac{m_p c^2}{800}} \]
Simplifying this gives:
\[ = \frac{\frac{1}{2} \cdot \frac{1}{100}}{\frac{1}{800}} = \frac{\frac{1}{2} \cdot 800}{100} = \frac{400}{100} = 4 \]

### Final Answer:
The ratio of the kinetic energy of the proton to that of the alpha particle is \( 4:1 \).

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