Question

α-particle, proton & electron have same kinetic energy. Select correct order of their de-Broglie wavelength.

• A. \(λ_e > λ_p > λ_α\)
• B. \(λ_α > λ_e > λ_p\)
• C. \(λ_p = λ_α = λ_e\)
• D. \(λ_p > λ_e > λ_α\)

Answer 1

The Correct Option is A

 To solve the question about the de-Broglie wavelength of particles having the same kinetic energy, we will use the de-Broglie wavelength formula: \(\lambda = \frac{h}{p}\), where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle.

Since the kinetic energy (KE) is the same for all particles, we can express kinetic energy as:

\(KE = \frac{1}{2} mv^2 = \frac{p^2}{2m}\)

Solving for momentum \(p\), we have:

\(p = \sqrt{2m \cdot KE}\)

Substituting for \(p\) in the de-Broglie equation, we get:

\(\lambda = \frac{h}{\sqrt{2m \cdot KE}}\)

Since \(KE\) is constant for all, the wavelength \(\lambda\) is inversely proportional to the square root of the mass \(m\) of the particles: \(\lambda \propto \frac{1}{\sqrt{m}}\)

Now, considering the masses:

• The mass of an electron \(m_e\) is the smallest.
• The mass of a proton \(m_p\) is larger than that of an electron.
• The mass of an alpha particle \(m_α\) (which consists of 2 protons and 2 neutrons) is much larger than that of a proton or electron.

Thus the de-Broglie wavelength will be in the order:

\(λ_e > λ_p > λ_α\)

This is because \(\lambda \propto \frac{1}{\sqrt{m}}\) implies that the lighter the particle, the longer the wavelength.

Hence, the correct answer is: \(λ_e > λ_p > λ_α\)