Question 103: Dual Nature of Matter and Radiation

Question

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $$25 \%$$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

Accepted Answer

We know that the de-Broglie wavelength $$\lambda$$ of a particle is given by:
$$\lambda = \frac{h}{p}$$
where:
$$h$$ is Planck's constant,
$$p$$ is the momentum of the particle.
For a photon (which has zero rest mass), its energy $$E$$ and momentum $$p$$ are related by the equation:
$$E = cp$$
and its de-Broglie wavelength $$\lambda$$ is given by:
$$\lambda = \frac{h}{E} 	imes \frac{1}{c}$$
Now, since the photon and electron are said to have the same de-Broglie wavelength:
$$\lambda_{electron} = \lambda_{photon}$$
$$\frac{h}{p_{electron}} = \frac{h}{E_{photon}} 	imes \frac{1}{c}$$
For the electron, its momentum $$p_{electron}$$ is given by:
$$p_{electron} = m_{e}v_{electron}$$
where:
$$m_{e}$$ is the electron's rest mass,
$$v_{electron}$$ is the electron's velocity.
The kinetic energy $$K.E.$$ of the electron is:
$$K.E._{electron} = \frac{1}{2}m_{e}v_{electron}^2$$
The question states that $$v_{electron}$$ is $$25\%$$ ($$0.25c$$) of the speed of light $$c$$. So we write:
$$v_{electron} = 0.25c$$
Plugging this into the kinetic energy formula, we get:
$$K.E._{electron} = \frac{1}{2}m_{e}(0.25c)^2 = \frac{1}{2}m_{e} 	imes \frac{1}{16}c^2$$
$$K.E._{electron} = \frac{1}{32}m_{e}c^2$$
For a photon, $$p_{photon} = \frac{E_{photon}}{c}$$ and hence its kinetic energy (which, for a photon, is simply its energy) is:
$$K.E._{photon} = E_{photon} = cp_{photon}$$
Now, comparing the kinetic energies:
$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{cp_{photon}}$$
Since $$p_{electron} = p_{photon}$$ (from the de-Broglie relation), we can replace $$p_{photon}$$ with $$p_{electron}$$ which is $$m_{e}v_{electron}$$:
$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{m_{e}v_{electron}c}$$
$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}c}{0.25c}$$
$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} 	imes \frac{1}{0.25}$$
$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} 	imes 4$$
$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{8}$$
So the correct answer is Option C $$\frac{1}{8}$$.

Answer

$$\frac{1}{4}$$

Answer

$$\frac{8}{1}$$

Answer

$$\frac{1}{8}$$

Answer

$$\frac{1}{1}$$

Dual Nature of Matter and Radiation

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $25 \%$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

Choose the correct answer: