Question

Kinetic energy of electron, proton and a particle is given as K, 2K and 4K respectively, then which of the following gives the correct order of De-Broglie wavelengths of electron, proton and a particle

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

Answer 1

The Correct Option is C

To determine the order of the De Broglie wavelengths of the electron, proton, and alpha particle, we start by recalling De Broglie's wavelength formula:

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

where \(h\) is Planck's constant and \(p\) is the momentum of the particle. The momentum can be expressed in terms of kinetic energy (\(K\)) as:

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

Substituting momentum in De Broglie's equation gives:

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

Thus, the De Broglie wavelength is inversely proportional to the square root of the product of the mass of the particle and its kinetic energy. Let's apply this to each given particle:

1. For the electron with kinetic energy \(K\): \(\lambda_e = \frac{h}{\sqrt{2m_eK}}\)
2. For the proton with kinetic energy \(2K\): \(\lambda_p = \frac{h}{\sqrt{2m_p(2K)}} = \frac{h}{\sqrt{4m_pK}} = \frac{h}{2\sqrt{m_pK}}\)
3. For the alpha particle with kinetic energy \(4K\): \(\lambda_\alpha = \frac{h}{\sqrt{2(4m_\alpha)K}} = \frac{h}{2\sqrt{2m_\alpha K}}\)

To compare wavelengths, consider the masses: the mass of an alpha particle (\(m_\alpha\)) is about 4 times the mass of a proton (\(m_p\)) and many times the mass of an electron (\(m_e\)).

Since the wavelength is inversely proportional to both the mass and the square root of the kinetic energy, and given all different kinetic energy values, we can establish:

1. \(\lambda_e\) is the greatest due to the smallest mass and single \(K\) factor.
2. \(\lambda_p\) is greater than \(\lambda_\alpha\) due to both the greater mass of an alpha particle and it being influenced by a higher \(K\) factor.

Thus, the correct order of De Broglie wavelengths is:

\(\lambda_e > \lambda_p > \lambda_\alpha\)