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)
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)
Solution and Explanation
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 \( \frac{1}{x} = \frac{1}{43} \).
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Concepts Used:
De Broglie Hypothesis
One of the equations that are commonly used to define the wave properties of matter is the de Broglie equation. Basically, it describes the wave nature of the electron.
De Broglie Equation Derivation and de Broglie Wavelength
Very low mass particles moving at a speed less than that of light behave like a particle and waves. De Broglie derived an expression relating to the mass of such smaller particles and their wavelength.
Plank’s quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.
E = hν …….(1)
E = mc2……..(2)
As the smaller particle exhibits dual nature, and energy being the same, de Broglie equated both these relations for the particle moving with velocity ‘v’ as,
This equation relating the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated using this relation is the de Broglie wavelength.