A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is `1.878 xx10^(-4) ` . The mass of the particle is close to :

Options

A. `4.8 xx 10^(-27)` kg
B. `9.1 xx10^(-31)` kg
C. `9.7 xx10^(-28)` kg
D. `1.2 xx10^(-28)` kg

Explanation

To solve the problem, we need to find the mass of a particle that is moving 5 times as fast as an electron, given the ratio of their de-Broglie wavelengths.

### Step-by-Step Solution:

1. **Understanding the de-Broglie wavelength**:
The de-Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is the Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

2. **Setting up the problem**:
Let:
- \( m_e \) = mass of the electron = \( 9.1 \times 10^{-31} \) kg
- \( v_e \) = velocity of the electron
- \( m_p \) = mass of the particle
- \( v_p = 5v_e \) (since the particle is moving 5 times as fast as the electron)

3. **Calculating the ratio of de-Broglie wavelengths**:
The ratio of the de-Broglie wavelengths of the particle and the electron is given by:
\[ \frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p v_p} \]
Substituting \( v_p = 5v_e \):
\[ \frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p (5v_e)} = \frac{m_e}{5m_p} \]

4. **Using the given ratio**:
We know from the problem that:
\[ \frac{\lambda_p}{\lambda_e} = 1.878 \times 10^{-4} \]
Therefore, we can set up the equation:
\[ 1.878 \times 10^{-4} = \frac{m_e}{5m_p} \]

5. **Rearranging to find \( m_p \)**:
Rearranging the equation gives:
\[ m_p = \frac{m_e}{5 \times 1.878 \times 10^{-4}} \]

6. **Substituting the value of \( m_e \)**:
Substituting \( m_e = 9.1 \times 10^{-31} \) kg:
\[ m_p = \frac{9.1 \times 10^{-31}}{5 \times 1.878 \times 10^{-4}} \]

7. **Calculating \( m_p \)**:
First, calculate \( 5 \times 1.878 \times 10^{-4} \):
\[ 5 \times 1.878 \times 10^{-4} = 9.39 \times 10^{-4} \]
Now, calculate \( m_p \):
\[ m_p = \frac{9.1 \times 10^{-31}}{9.39 \times 10^{-4}} \approx 9.7 \times 10^{-28} \text{ kg} \]

### Final Answer:
The mass of the particle is approximately \( 9.7 \times 10^{-28} \) kg.