An electron travels with a velocity of x `ms^(-1).` For a proton to have the same de-Broglie wavelength, the velocity will be approximately:

Options

A. `(1840)/(x)`
B. `(x)/(1840)`
C. `1840 " x "`
D. x

Answer

To find the velocity of a proton that has the same de-Broglie wavelength as an electron traveling at a velocity of \( x \, \text{ms}^{-1} \), we can follow these steps:

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

### Step 2: Calculate the de-Broglie wavelength for the electron
For the electron, the de-Broglie wavelength (\( \lambda_e \)) can be expressed as:
\[ \lambda_e = \frac{h}{m_e \cdot x} \]
where \( m_e \) is the mass of the electron and \( x \) is its velocity.

### Step 3: Write the de-Broglie wavelength for the proton
For the proton, the de-Broglie wavelength (\( \lambda_p \)) is:
\[ \lambda_p = \frac{h}{m_p \cdot v_p} \]
where \( m_p \) is the mass of the proton and \( v_p \) is its velocity.

### Step 4: Set the wavelengths equal
Since we want the proton to have the same de-Broglie wavelength as the electron, we set \( \lambda_e = \lambda_p \):
\[ \frac{h}{m_e \cdot x} = \frac{h}{m_p \cdot v_p} \]

### Step 5: Cancel out Planck's constant
We can cancel \( h \) from both sides:
\[ \frac{1}{m_e \cdot x} = \frac{1}{m_p \cdot v_p} \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ m_e \cdot x = m_p \cdot v_p \]

### Step 7: Substitute the mass of the proton
The mass of the proton (\( m_p \)) is approximately 1840 times the mass of the electron (\( m_e \)):
\[ m_e \cdot x = (1840 \cdot m_e) \cdot v_p \]

### Step 8: Cancel the mass of the electron
Dividing both sides by \( m_e \):
\[ x = 1840 \cdot v_p \]

### Step 9: Solve for the velocity of the proton
Now, we can solve for \( v_p \):
\[ v_p = \frac{x}{1840} \]

### Conclusion
Thus, the velocity of the proton to have the same de-Broglie wavelength as the electron is:
\[ v_p \approx \frac{x}{1840} \]