A particle is moving three times as fast as an electron. The ratio of the de- Broglie wavelength of the particle to that of the electron is `1.813xx10^-4`. Calculate the particle's mass and identify the particle. Mass of electron `=9.11xx10^(-31)kg`.

Options

A. `1.67xx10^(-27)kg`
B. `1.67xx10^(-31)kg`
C. `1.67xx10^(-19)kg`
D. `1.67xx10^(-14)kg`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.

### Step 2: Write the expression for the ratio of wavelengths
Given that the particle is moving three times as fast as the electron, we can denote:
- \( v_e \) as the velocity of the electron,
- \( v = 3v_e \) as the velocity of the particle.

The ratio of the de Broglie wavelengths of the particle and the electron can be expressed as:
\[ \frac{\lambda}{\lambda_e} = \frac{m_e v_e}{mv} \]
Substituting \( v = 3v_e \):
\[ \frac{\lambda}{\lambda_e} = \frac{m_e v_e}{m(3v_e)} = \frac{m_e}{3m} \]

### Step 3: Use the given ratio of wavelengths
From the problem, we know:
\[ \frac{\lambda}{\lambda_e} = 1.813 \times 10^{-4} \]
Thus, we can write:
\[ \frac{m_e}{3m} = 1.813 \times 10^{-4} \]

### Step 4: Solve for the mass of the particle
Rearranging the equation gives:
\[ m = \frac{m_e}{3 \times 1.813 \times 10^{-4}} \]
Substituting the mass of the electron \( m_e = 9.11 \times 10^{-31} \, \text{kg} \):
\[ m = \frac{9.11 \times 10^{-31}}{3 \times 1.813 \times 10^{-4}} \]

### Step 5: Calculate the mass
Calculating the denominator:
\[ 3 \times 1.813 \times 10^{-4} = 5.439 \times 10^{-4} \]
Now, substituting back:
\[ m = \frac{9.11 \times 10^{-31}}{5.439 \times 10^{-4}} \approx 1.67 \times 10^{-27} \, \text{kg} \]

### Step 6: Identify the particle
The mass \( 1.67 \times 10^{-27} \, \text{kg} \) is approximately the mass of a proton, which is around \( 1.67 \times 10^{-27} \, \text{kg} \). Thus, the particle is likely a proton.

### Final Answer
The mass of the particle is approximately \( 1.67 \times 10^{-27} \, \text{kg} \), and the particle is identified as a proton.

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