Linear momenta of a proton and an electrons are euqal. Relative to an electron-

Options

A. Kinetic energy of proton is more
B. De-Brogile wavelength of proton is more.
C. De-Brogile wavelength of proton is less
D. De-Brogile wavelength of proton and electron are equal

Explanation

To solve the problem, we need to analyze the situation where the linear momenta of a proton and an electron are equal. We will derive the implications of this condition step by step.

### Step 1: Understand the concept of linear momentum
Linear momentum (p) is defined as the product of mass (m) and velocity (v):
\[ p = mv \]
For a proton and an electron, we have:
\[ p_{\text{proton}} = m_{\text{proton}} \cdot v_{\text{proton}} \]
\[ p_{\text{electron}} = m_{\text{electron}} \cdot v_{\text{electron}} \]

### Step 2: Set the momenta equal
According to the problem, the linear momenta of the proton and the electron are equal:
\[ m_{\text{proton}} \cdot v_{\text{proton}} = m_{\text{electron}} \cdot v_{\text{electron}} \]

### Step 3: Relate the velocities
From the equation above, we can express the velocity of the proton in terms of the velocity of the electron:
\[ v_{\text{proton}} = \frac{m_{\text{electron}}}{m_{\text{proton}}} \cdot v_{\text{electron}} \]

### Step 4: Calculate the de Broglie wavelength
The de Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant.

Since the momenta are equal, we can write:
\[ \lambda_{\text{proton}} = \frac{h}{p_{\text{proton}}} \]
\[ \lambda_{\text{electron}} = \frac{h}{p_{\text{electron}}} \]

Since \(p_{\text{proton}} = p_{\text{electron}}\), it follows that:
\[ \lambda_{\text{proton}} = \lambda_{\text{electron}} \]

### Step 5: Analyze kinetic energy
The kinetic energy (K.E.) of an object is given by:
\[ K.E. = \frac{1}{2} mv^2 \]

For the proton and electron, we can write:
\[ K.E._{\text{proton}} = \frac{1}{2} m_{\text{proton}} v_{\text{proton}}^2 \]
\[ K.E._{\text{electron}} = \frac{1}{2} m_{\text{electron}} v_{\text{electron}}^2 \]

### Step 6: Substitute for velocities
Using the relationship from Step 3, we can substitute \(v_{\text{proton}}\):
\[ K.E._{\text{proton}} = \frac{1}{2} m_{\text{proton}} \left(\frac{m_{\text{electron}}}{m_{\text{proton}}} v_{\text{electron}}\right)^2 \]
\[ = \frac{1}{2} \frac{m_{\text{electron}}^2}{m_{\text{proton}}} v_{\text{electron}}^2 \]

### Step 7: Compare kinetic energies
Now, we can compare the kinetic energies:
Since \(m_{\text{proton}} > m_{\text{electron}}\), it follows that:
\[ K.E._{\text{proton}} < K.E._{\text{electron}} \]

### Conclusion
Thus, while the de Broglie wavelengths of the proton and electron are equal, the kinetic energy of the proton is less than that of the electron.

### Final Answer
The correct conclusion is that the kinetic energy of the proton is less than that of the electron.

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