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 find the ratio of kinetic energies of a proton and an electron given they have the same de-Broglie wavelength, we need to follow several steps. The de-Broglie wavelength \( \lambda \) is given by the formula:

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

where \( h \) is Planck's constant and \( p \) is the momentum. Since both the proton and electron have the same \( \lambda \), their momenta are equal:

\[p_e = p_p\]

The momentum \( p \) can also be expressed as:

\[p = \sqrt{2mK}\]

where \( m \) is the mass and \( K \) is the kinetic energy. Setting the momenta of the electron and proton equal gives:

\[\sqrt{2m_eK_e} = \sqrt{2m_pK_p}\]

Squaring both sides, we get:

\[2m_eK_e = 2m_pK_p\]

Simplifying, we have:

\[m_eK_e = m_pK_p\]

From this, the ratio of kinetic energies \( K_e \) and \( K_p \) is:

\[\frac{K_e}{K_p} = \frac{m_p}{m_e}\]

Given that the mass of the proton \( m_p \) is 1836 times the mass of the electron \( m_e \), we substitute:

\[\frac{K_e}{K_p} = 1836\]

Thus, the ratio of the kinetic energies of the electron to the proton is given by:

\[1 : 1836\]

This confirms that the correct answer is 1 : 1836.

Answer 2

For the same de-Broglie wavelength, \(P = \frac{h}{\lambda}\) is the same for both the proton and the electron. Kinetic energy is given by:
\[ \text{KE} = \frac{P^2}{2m}. \]
Thus:
\[ \frac{\text{KE}_e}{\text{KE}_p} = \frac{m_p}{m_e}. \]
Given:
\[ m_p = 1836 m_e. \]
Substitute:
\[ \frac{\text{KE}_e}{\text{KE}_p} = \frac{1}{1836}. \]
Final Answer: \(1 : 1836\).