Question 63: Dual Nature of Matter and Radiation

Question

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :

Accepted Answer

To determine the relationship between the de-Broglie wavelengths of a proton, an electron, and an alpha particle with the same energies, we need to use the de-Broglie wavelength formula:
$$\lambda = \frac{h}{p}$$
where $ h $ is Planck's constant and $ p $ is the momentum of the particle.
For particles with the same kinetic energy $ E $, we have:
$$E = \frac{p^2}{2m}$$
Solving for $ p $:
$$p = \sqrt{2mE}$$
Substituting this into the de-Broglie equation, we get:
$$\lambda = \frac{h}{\sqrt{2mE}}$$
Since all three particles have the same energy $ E $, the de-Broglie wavelength is inversely proportional to the square root of the mass $ m $:
$$\lambda \propto \frac{1}{\sqrt{m}}$$
The masses of the particles are as follows:
Mass of proton $ m_p $ is approximately $ 1.67 	imes 10^{-27} $ kg
Mass of electron $ m_e $ is approximately $ 9.11 	imes 10^{-31} $ kg
Mass of alpha particle $ m_{\alpha} $ is approximately $ 4 	imes 1.67 	imes 10^{-27} $ kg (since it has 2 protons and 2 neutrons)
Comparatively:
Mass of alpha particle $ m_{\alpha} $ is the largest.
Mass of proton $ m_p $ is intermediate.
Mass of electron $ m_e $ is the smallest.
Therefore, the de-Broglie wavelength will be:
Largest for the electron ($ \lambda_{\mathrm{e}} $)
Intermediate for the proton ($ \lambda_{\mathrm{p}} $)
Smallest for the alpha particle ($ \lambda_{\alpha} $)
Hence, the correct order is:
Option B:
$$\lambda_\alpha<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}$$

Answer

$$\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_\alpha$$

Answer

$$\lambda_\alpha<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}$$

Answer

$$\lambda_{\mathrm{e}}>\lambda_\alpha>\lambda_{\mathrm{p}}$$

Answer

$$\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}<\lambda_\alpha$$

Dual Nature of Matter and Radiation

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :

Choose the correct answer: