JEE Chemistry 2023 Question with Solution

Given: An $\alpha$-particle, a proton and an electron have the same kinetic energy.

Find: The correct order of their de-Broglie wavelengths.

The de-Broglie wavelength is given by

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$

For particles with the same kinetic energy,

$$\lambda \propto \frac{1}{\sqrt{m}}$$

Since

$$m_{\alpha} > m_p > m_e$$

therefore,

$$\lambda_e > \lambda_p > \lambda_{\alpha}$$

This is equivalent to

$$\lambda_{\alpha} < \lambda_p < \lambda_e$$

the solution states the correct ordering corresponds to option A, but with the given options this ordering is actually option B. Therefore, the correct option is B.

Given: Same kinetic energy for electron, proton and $\alpha$-particle.

Find: Which particle has the largest and smallest de-Broglie wavelength.

Using

$$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}$$

when kinetic energy is the same, wavelength depends only on mass as

$$\lambda \propto \frac{1}{\sqrt{m}}$$

The masses listed in the solution are:

• electron $= 9.1 \times 10^{-31} \, \text{kg}$
• proton $= 1.67 \times 10^{-27} \, \text{kg}$
• $\alpha$-particle $= 6.68 \times 10^{-27} \, \text{kg}$

So the lightest particle, the electron, has the greatest wavelength, and the heaviest particle, the $\alpha$-particle, has the smallest wavelength. Thus,

$$\lambda_e > \lambda_p > \lambda_{\alpha}$$

Hence in the option format,

$$\lambda_{\alpha} < \lambda_p < \lambda_e$$

So the correct option is B.