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

To compare the de-Broglie wavelengths of particles with the same energies, we need to understand the de-Broglie wavelength formula and how it is affected by the mass and energy of the particles.

The de-Broglie wavelength \(\lambda\) is given by the formula:

\(\lambda = \dfrac{h}{p}\)

where:

• \(h\) is the Planck's constant.
• \(p\) is the momentum of the particle.

For a particle with kinetic energy \(E\), the momentum can be expressed as:

\(p = \sqrt{2mE}\)

Substituting this in the de-Broglie equation, we get:

\(\lambda = \dfrac{h}{\sqrt{2mE}}\)

From this equation, we see that for particles with the same energy \(E\), the wavelength \(\lambda\) is inversely proportional to the square root of their masses:

\(\lambda \propto \dfrac{1}{\sqrt{m}}\)

Now, let's compare the masses of the given particles:

• Proton (\(m_p\)): Approximately \(1.67 \times 10^{-27}\) kg
• Electron (\(m_e\)): Approximately \(9.11 \times 10^{-31}\) kg
• Alpha particle (\(m_\alpha\)): Consists of 2 protons and 2 neutrons, approximately \(4 \times 1.67 \times 10^{-27}\) kg = \(6.68 \times 10^{-27}\) kg

Based on these masses:

• The electron is the lightest, hence it has the longest wavelength.
• The proton is heavier than the electron, but lighter than the alpha particle, so it has an intermediate wavelength.
• The alpha particle is the heaviest, hence it has the shortest wavelength.

Therefore, the correct order of the de-Broglie wavelengths is:

\(\lambda_\alpha < \lambda_p < \lambda_e\)

Thus, the correct answer is:

\(\lambda_\alpha<\lambda_p<\lambda_e\)

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