Question

A proton, an electron, and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:

• A. \(\lambda_e>\lambda_\alpha>\lambda_p\)
• B. \(\lambda_\alpha<\lambda_p<\lambda_e\)
• C. \(\lambda_p<\lambda_e<\lambda_\alpha\)
• D. \(\lambda_p>\lambda_e>\lambda_\alpha\)

Answer 1

The Correct Option is B

The de-Broglie wavelength is given by:

\[ \lambda_{DB} = \frac{h}{p} = \frac{h}{\sqrt{2mK}}, \]

where \(m\) is the mass, \(K\) is the kinetic energy, and \(p\) is the momentum. Since the particles have the same energy \(K\), the de-Broglie wavelength becomes:

\[ \lambda_{DB} \propto \frac{1}{\sqrt{m}}. \]

Step 1: Compare masses

• Electron (\(m_e\)): Lightest particle.
• Proton (\(m_p\)): Heavier than an electron.
• Alpha particle (\(m_\alpha\)): Heaviest, \(m_\alpha = 4m_p\).

Step 2: Compare wavelengths

Since \(\lambda_{DB} \propto \frac{1}{\sqrt{m}}\), the lighter the mass, the longer the wavelength:

\[ \lambda_e > \lambda_p > \lambda_\alpha. \]

Final Answer:
\[ \lambda_\alpha < \lambda_p < \lambda_e. \]