Question

Proton and electron have equal kinetic energy, the ratio of de-Broglie wavelength of proton and electron is \(\frac{1}{x}\). Find x. (Mass of proton 1849 times mass of electron)

Answer 1

Step 1: Formula for de Broglie Wavelength 

The de Broglie wavelength (\( \lambda \)) of a particle is given by:

\[ \lambda = \frac{h}{p}, \]

where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

Step 2: Momentum and Kinetic Energy

The momentum of a particle is expressed as:

\[ p = mv, \]

where \( m \) is the mass of the particle and \( v \) is its velocity. From the expression for kinetic energy \( K = \frac{1}{2}mv^2 \), the velocity can be written as:

• For the proton: \[ v_p = \sqrt{\frac{2K}{m_p}}. \]
• For the electron: \[ v_e = \sqrt{\frac{2K}{m_e}}. \]

Step 3: Ratio of de Broglie Wavelengths

The ratio of the de Broglie wavelengths of the proton and the electron is given by:

\[ \frac{\lambda_p}{\lambda_e} = \frac{p_e}{p_p} = \frac{m_e v_e}{m_p v_p}. \]

Substituting the expressions for \( v_p \) and \( v_e \):

\[ \frac{\lambda_p}{\lambda_e} = \frac{m_e \cdot \sqrt{\frac{2K}{m_e}}}{m_p \cdot \sqrt{\frac{2K}{m_p}}}. \]

Simplify the expression:

\[ \frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{m_p}}. \]

Step 4: Mass Ratio

Given that the mass of the proton (\( m_p \)) is 1849 times the mass of the electron (\( m_e \)), we have:

\[ \frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{1849 m_e}} = \frac{1}{43}. \]

Final Answer:

The value of \( x \) is 43, and the ratio of de Broglie wavelengths is \( \frac{1}{x} = \frac{1}{43} \).