Question 85: Dual Nature of Matter and Radiation

Question

A proton and an electron have the same de Broglie wavelength. If $$\mathrm{K}_{\mathrm{p}}$$ and $$\mathrm{K}_{\mathrm{e}}$$ be the kinetic energies of proton and electron respectively, then choose the correct relation :

Accepted Answer

To determine the correct relation between the kinetic energies of a proton ($$\mathrm{K}_{\mathrm{p}}$$) and an electron ($$\mathrm{K}_{\mathrm{e}}$$) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength 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 $$p$$ of a particle is given by:
$$p = \sqrt{2mK}$$
where $$m$$ is the mass of the particle and $$K$$ is its kinetic energy. For a proton and an electron with the same de Broglie wavelength:
$$\lambda_{\mathrm{p}} = \lambda_{\mathrm{e}}$$
This implies the momenta should be the same:
$$p_{\mathrm{p}} = p_{\mathrm{e}}$$
Thus, we can write:
$$\sqrt{2 m_{\mathrm{p}} K_{\mathrm{p}}} = \sqrt{2 m_{\mathrm{e}} K_{\mathrm{e}}}$$
Squaring both sides to eliminate the square roots:
$$2 m_{\mathrm{p}} K_{\mathrm{p}} = 2 m_{\mathrm{e}} K_{\mathrm{e}}$$
$$m_{\mathrm{p}} K_{\mathrm{p}} = m_{\mathrm{e}} K_{\mathrm{e}}$$
Rearranging to solve for $$K_{\mathrm{p}}$$ in terms of $$K_{\mathrm{e}}$$:
$$K_{\mathrm{p}} = \frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} K_{\mathrm{e}}$$
Since the mass of a proton $$m_{\mathrm{p}}$$ is much greater than the mass of an electron $$m_{\mathrm{e}}$$:
$$m_{\mathrm{p}} \gg m_{\mathrm{e}}$$
This means:
$$\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} \ll 1$$
Therefore, it implies:
$$K_{\mathrm{p}} < K_{\mathrm{e}}$$
So, the correct option is:
Option C: $$\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$$

Answer

$$\mathrm{K_p>K_e}$$

Answer

$$\mathrm{K_p=K_e}$$

Answer

$$\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$$

Answer

$$\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}{ }^2$$

Dual Nature of Matter and Radiation

A proton and an electron have the same de Broglie wavelength. If $\mathrm{K}_{\mathrm{p}} and \mathrm{K}_{\mathrm{e}}$ be the kinetic energies of proton and electron respectively, then choose the correct relation :

Choose the correct answer: