Question

A proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
Given: \(h = 6.63 \times 10^{-34} \, \text{Js}\), \(m_e = 9.0 \times 10^{-31} \, \text{kg}\), and \(m_p = 1836 \times m_e\)

• A. \(1 : 1836\)
• B. \(1 : \frac{1}{1836}\)
• C. \(1 : \frac{1}{\sqrt{1836}}\)
• D. \(1 : \sqrt{1836}\)

Answer 1

The Correct Option is A

To determine the ratio of the kinetic energies of a proton and an electron with identical de-Broglie wavelengths, follow these steps. The de-Broglie wavelength \( \lambda \) is defined as \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is momentum. Equal wavelengths imply equal momenta: \( p_e = p_p \). Momentum can also be expressed as \( p = \sqrt{2mK} \), where \( m \) is mass and \( K \) is kinetic energy. Equating the momenta yields \( \sqrt{2m_eK_e} = \sqrt{2m_pK_p} \). Squaring both sides gives \( 2m_eK_e = 2m_pK_p \), which simplifies to \( m_eK_e = m_pK_p \). Therefore, the ratio of kinetic energies \( K_e \) to \( K_p \) is \( \frac{K_e}{K_p} = \frac{m_p}{m_e} \). Given that a proton's mass \( m_p \) is 1836 times an electron's mass \( m_e \), we have \( \frac{K_e}{K_p} = 1836 \). The ratio of the electron's kinetic energy to the proton's kinetic energy is thus \( 1 : 1836 \).