Question

The kinetic energy of an electron, α-particle and a proton are given as 4K, 2K and K respectively. The de-Broglie wavelength associated with electron (λe) α-particle (λα) and the proton (λp) are as follows :

• A. λα > λp > λe
• B. λα = λp > λe
• C. λα = λp < λe
• D. λα < λp < λe

Answer 1

The Correct Option is D

 To determine the de-Broglie wavelengths of the electron, α-particle, and proton, we have to understand the relationship between their kinetic energies and their wavelengths.

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

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

Where \(h\) is Planck's constant, and \(p\) is momentum. Momentum \(p\) can also be expressed in terms of kinetic energy \(K\) as:

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

Substituting in the de-Broglie equation gives:

\(\lambda = \frac{h}{\sqrt{2mK}}\)

Thus, \(\lambda \propto \frac{1}{\sqrt{K}}\) for a given particle.

Now, given the kinetic energies:

• Electron: \(4K\)
• α-particle: \(2K\)
• Proton: \(K\)

The masses of these particles are different and must be considered:

• Electron mass \(m_e\)
• α-particle mass \(m_{\alpha}<\span>\)
• Proton mass \(m_p\)

The de-Broglie wavelengths can be compared as:

\(\lambda_e = \frac{h}{\sqrt{2m_e \cdot 4K}} = \frac{h}{2 \sqrt{2m_e K}}\)

\(\lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha} \cdot 2K}} = \frac{h}{\sqrt{4m_{\alpha} K}}\)

\(\lambda_p = \frac{h}{\sqrt{2m_p \cdot K}}\)

Since \(\lambda \propto \frac{1}{\sqrt{K}}\), the particle with the lowest mass and kinetic energy will have the longest de-Broglie wavelength.

Comparing wavelengths:

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

Thus, the de-Broglie wavelengths sorted from shortest to longest are: \(λα < λp < λe\). Therefore, the correct answer is:

λα < λp < λe