An electron and a proton have same kinetic energy Ratio of their respective de-Broglie wavelength is about

Options

A. `(12.27)/(0.266)`
B. `(0.101)/(0.266) `
C. `(0.266)/(12.27)`
D. `(0.101)/(12.27)`

Answer

To solve the problem of finding the ratio of the de-Broglie wavelengths of an electron and a proton with the same kinetic energy, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

### Step 2: Express momentum in terms of kinetic energy
The momentum \(p\) of a particle can be expressed in terms of its kinetic energy (K) and mass (m) as:
\[ p = \sqrt{2Km} \]
Thus, we can rewrite the de-Broglie wavelength as:
\[ \lambda = \frac{h}{\sqrt{2Km}} \]

### Step 3: Set up the ratio of wavelengths for the electron and proton
Let \( \lambda_e \) be the wavelength of the electron and \( \lambda_p \) be the wavelength of the proton. Since both particles have the same kinetic energy, we can write:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2K m_p}}{\sqrt{2K m_e}} = \frac{1}{\sqrt{\frac{m_p}{m_e}}} \]
This simplifies to:
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_e}{m_p}} \]

### Step 4: Substitute the masses of the electron and proton
The mass of the electron \(m_e\) is approximately \(9.1 \times 10^{-31} \, \text{kg}\) and the mass of the proton \(m_p\) is approximately \(1.67 \times 10^{-27} \, \text{kg}\). Now we can calculate the ratio:
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{9.1 \times 10^{-31}}{1.67 \times 10^{-27}}} \]

### Step 5: Calculate the ratio
Calculating the mass ratio:
\[ \frac{9.1 \times 10^{-31}}{1.67 \times 10^{-27}} \approx 5.45 \times 10^{-4} \]
Taking the square root:
\[ \sqrt{5.45 \times 10^{-4}} \approx 0.0233 \]
Thus, the ratio of the wavelengths is:
\[ \frac{\lambda_e}{\lambda_p} \approx 0.0233 \]

### Step 6: Find the inverse ratio
Since we want the ratio of the wavelengths in the form of \( \frac{\lambda_e}{\lambda_p} \), we can also express it as:
\[ \frac{\lambda_p}{\lambda_e} = \frac{1}{0.0233} \approx 42.9 \]

### Conclusion
The ratio of the de-Broglie wavelengths of the electron to the proton is approximately 43.