Proton (P) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, `m_(p)=1849 m_(c)`)

Options

A. `1:1`
B. `1:1849`
C. `1:43`
D. `43:1`

Answer

To solve the problem, we need to find the ratio of the momenta of a proton and an electron when they have the same de Broglie wavelength.

### Step-by-Step Solution:

1. **Understanding de Broglie Wavelength**:
The de Broglie wavelength (\( \lambda \)) 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.

2. **Setting Up the Equation**:
Since we are given that the de Broglie wavelengths of the proton and the electron are the same, we can write:
\[ \lambda_p = \lambda_e \]
This implies:
\[ \frac{h}{p_p} = \frac{h}{p_e} \]
where \( p_p \) is the momentum of the proton and \( p_e \) is the momentum of the electron.

3. **Cancelling Planck's Constant**:
We can cancel \( h \) from both sides of the equation:
\[ \frac{1}{p_p} = \frac{1}{p_e} \]
This leads to:
\[ p_p = p_e \]

4. **Expressing Momentum in Terms of Mass and Velocity**:
The momentum of a particle is given by:
\[ p = mv \]
Therefore, we can express the momenta of the proton and electron as:
\[ p_p = m_p v_p \quad \text{and} \quad p_e = m_e v_e \]
where \( m_p \) and \( m_e \) are the masses of the proton and electron, respectively, and \( v_p \) and \( v_e \) are their velocities.

5. **Using the Given Mass Ratio**:
We are given that the mass of the proton is \( m_p = 1849 m_e \). Substituting this into our momentum equation gives:
\[ 1849 m_e v_p = m_e v_e \]

6. **Simplifying the Equation**:
Dividing both sides by \( m_e \) (assuming \( m_e \neq 0 \)):
\[ 1849 v_p = v_e \]

7. **Finding the Ratio of Momenta**:
Now, we can find the ratio of the momenta:
\[ \frac{p_p}{p_e} = \frac{m_p v_p}{m_e v_e} = \frac{1849 m_e v_p}{m_e (1849 v_p)} = \frac{1849}{1} \]

8. **Final Result**:
Thus, the ratio of their momenta is:
\[ \frac{p_p}{p_e} = 1849 : 1 \]

### Conclusion:
The ratio of the momentum of the proton to that of the electron when they have the same de Broglie wavelength is \( 1849 : 1 \).