Question 116: Dual Nature of Matter and Radiation

Question

The kinetic energy of an electron, $$\alpha$$-particle and a proton are given as $$4 \mathrm{~K}, 2 \mathrm{~K}$$ and $$\mathrm{K}$$ respectively. The de-Broglie wavelength associated with electron $$(\lambda \mathrm{e}), \alpha$$-particle $$((\lambda \alpha)$$ and the proton $$(\lambda p)$$ are as follows:

Accepted Answer

The de Broglie wavelength of a particle is given by the equation:
$ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}, $
where:
$h$ is Planck's constant,
$p$ is the momentum of the particle,
$m$ is the mass of the particle, and
$K$ is the kinetic energy of the particle.
This equation shows that the de Broglie wavelength of a particle is inversely proportional to the square root of its mass and its kinetic energy. This means that the particle with the smallest mass and kinetic energy will have the largest de Broglie wavelength.
Given that the kinetic energies of the electron, alpha particle, and proton are $4K$, $2K$, and $K$, respectively, and knowing that the mass of the electron is less than the mass of the proton and the mass of the proton is less than the mass of the alpha particle, we can infer that the de Broglie wavelength of the electron is greater than the de Broglie wavelength of the proton, which in turn is greater than the de Broglie wavelength of the alpha particle.
Therefore, the correct answer is $\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$.

Answer

$$\lambda \alpha<\lambda p<\lambda e$$

Answer

$$\lambda \alpha>\lambda p>\lambda e$$

Answer

$$\lambda \alpha=\lambda p<\lambda e$$

Answer

$$\lambda \alpha=\lambda p>\lambda e$$

Dual Nature of Matter and Radiation

The kinetic energy of an electron, $\alpha-particle and a proton are given as 4 \mathrm{~K}, 2 \mathrm{~K} and \mathrm{K} respectively. The de-Broglie wavelength associated with electron (\lambda \mathrm{e}), \alpha-particle ((\lambda \alpha) and the proton (\lambda p)$ are as follows:

Choose the correct answer: