Question 261: Dual Nature of Matter and Radiation

Question

Let $E_c$ and $E_p$ represents kinetic energy of electron and photon respectively. If de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron then $E_p / E_c$ is (speed of electron $=\mathrm{C} / 100$ where C is the velocity of light)

Accepted Answer

Step 1: Write the energy of the photon
The energy of a photon is given by: $$E_p = h
u = \frac{hc}{\lambda_p}$$
We know from the question that the photonÃ¢ÂÂs de-Broglie wavelength ($\lambda_p$) is twice the electronÃ¢ÂÂs de-Broglie wavelength ($\lambda_e$): $$\lambda_p = 2\lambda_e$$
So, we substitute $\lambda_p$ into the energy formula: $$E_p = \frac{hc}{2\lambda_e}$$
Step 2: Write the kinetic energy of the electron
The kinetic energy of the electron is: $$E_c = \frac{p^2}{2m}$$ where $p$ is momentum and $m$ is the mass of the electron.
The momentum $p$ can also be written as: $$p = \frac{h}{\lambda_e}$$
Substitute this into the kinetic energy equation: $$E_c = \frac{1}{2m} \left(\frac{h}{\lambda_e}ight)^2 = \frac{h^2}{2m\lambda_e^2}$$
Step 3: Find the ratio $\frac{E_p}{E_c}$
Now substitute the formulas from Step 1 and Step 2:
$$\frac{E_p}{E_c} = \frac{\frac{hc}{2\lambda_e}}{\frac{h^2}{2m\lambda_e^2}}$$
The denominators are complex, but if we multiply by the reciprocal, it looks like this:
$$\frac{E_p}{E_c} = \frac{hc}{2\lambda_e} 	imes \frac{2m\lambda_e^2}{h^2}$$
Simplify the terms:
$$\frac{E_p}{E_c} = \frac{m c \lambda_e}{h}$$
Step 4: Replace $\lambda_e$ using speed of electron
The de-Broglie wavelength of electron is: $$\lambda_e = \frac{h}{p}$$ where $p$ is the electronÃ¢ÂÂs momentum.
Since the electronÃ¢ÂÂs speed $v = \frac{c}{100}$, then $p = m 	imes v = m 	imes \frac{c}{100}$.
So, $$\lambda_e = \frac{h}{m 	imes \frac{c}{100}} = \frac{100 h}{m c}$$
Substitute this back into the ratio:
$$\frac{E_p}{E_c} = \frac{m c}{h} 	imes \frac{100 h}{m c}$$
The $m$, $c$, and $h$ cancel out, so:
$$\frac{E_p}{E_c} = 100$$

Answer

10

Answer

$10^2$

Answer

$10^3$

Answer

$10^4$

Dual Nature of Matter and Radiation

Let E_c and E_p represents kinetic energy of electron and photon respectively. If de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron then E_p / E_c is (speed of electron =\mathrm{C} / 100 where C is the velocity of light)

Choose the correct answer: