Question

An \( \alpha \)-particle, a proton, and an electron have the same kinetic energy. Which one of the following is correct in case of their De-Broglie wavelength:

• A. \( \lambda_{\alpha} < \lambda_{p} < \lambda_{e} \)
• B. \( \lambda_{\alpha} > \lambda_{p} > \lambda_{e} \)
• C. \( \lambda_{\alpha} = \lambda_{p} = \lambda_{e} \)
• D. \( \lambda_{\alpha} > \lambda_{p} < \lambda_{e} \)

Answer 1

The Correct Option is A

To solve the given problem, let's first understand the relationship between De-Broglie wavelength and kinetic energy: 

The De-Broglie wavelength \(\lambda\) of a particle is given by the formula:

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

where \(\lambda\) is the De-Broglie wavelength, \(h\) is Planck's constant, and \(p\) is the momentum of the particle. The momentum \()\) can be expressed in terms of kinetic energy \((K)\) a\)

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

Thus, substituting for \(p\), we have:

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

Given that the kinetic energy \(K\) is the same for all particles (an \(\alpha\)-particle, a proton, and an electron), the De-Broglie wavelength depends on the mass \(m\) of the particle.

Since the mass \(m\) appears in the denominator, a larger mass results in a shorter wavelength.

Let's compare the masses of the particles:

• Mass of an \(\alpha\)-particle: Approximately 4 u (atomic mass units)
• Mass of a proton: Approximately 1 u
• Mass of an electron: Approximately \(1/1836\) u

Since the \(\alpha\)-particle has the largest mass, it will have the smallest De-Broglie wavelength. Conversely, the electron, with the smallest mass, will have the largest De-Broglie wavelength.

Thus, the order of wavelengths is: \(\lambda_{\alpha} < \lambda_{p} < \lambda_{e}\).

Therefore, the correct option is:

\((\lambda_{\alpha} < \lambda_{p} < \lambda_{e})\)