Question 86: Dual Nature of Matter and Radiation

Question

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume h = 6.63 $$	imes 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 	imes 10^{-31} \mathrm{~kg}$$ and $$\mathrm{m}_{\mathrm{p}}=1836$$ times $$\mathrm{m}_{\mathrm{e}}$$ )

Accepted Answer

To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.
The de-Broglie wavelength $$\lambda$$ is given by:
$$\lambda = \frac{h}{p}$$
where $$h$$ is PlanckÃ¢ÂÂs constant and $$p$$ is the momentum.
Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:
$$\lambda_{	ext{electron}} = \lambda_{	ext{proton}} \implies \frac{h}{p_{	ext{e}}} = \frac{h}{p_{	ext{p}}} \implies p_{	ext{e}} = p_{	ext{p}}$$
Next, the kinetic energy $$K$$ of a particle is related to its momentum $$p$$ and mass $$m$$ by the following formula:
$$K = \frac{p^2}{2m}$$
Given that the momentum $$p$$ is the same for both the proton and the electron, we can write the kinetic energies as:
$$K_{	ext{e}} = \frac{p^2}{2m_{	ext{e}}}$$
$$K_{	ext{p}} = \frac{p^2}{2m_{	ext{p}}}$$
The ratio of the kinetic energies is therefore:
$$\frac{K_{	ext{e}}}{K_{	ext{p}}} = \frac{\frac{p^2}{2m_{	ext{e}}}}{\frac{p^2}{2m_{	ext{p}}}} = \frac{m_{	ext{p}}}{m_{	ext{e}}}$$
Given that the mass of the proton $$m_{	ext{p}}$$ is 1836 times the mass of the electron $$m_{	ext{e}}$$, we have:
$$\frac{m_{	ext{p}}}{m_{	ext{e}}} = 1836$$
Thus, the ratio of the kinetic energies is:
$$\frac{K_{	ext{e}}}{K_{	ext{p}}} = 1836$$
Therefore, the correct answer is:
Option C
$$1: 1836$$

Answer

$$1: \frac{1}{1836}$$

Answer

$$1: \sqrt{1836}$$

Answer

$$1: 1836$$

Answer

$$1: \frac{1}{\sqrt{1836}}$$

Dual Nature of Matter and Radiation

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h = 6.63 $\times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg} and \mathrm{m}_{\mathrm{p}}=1836 times \mathrm{m}_{\mathrm{e}}$ )

Choose the correct answer: