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

Answer: (B)

$\lambda_\alpha<\lambda_p<\lambda_e$

Answer 2

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:

$\lambda_\alpha < \lambda_p < \lambda_e.$