Question 353: Dual Nature of Matter and Radiation

Question

An electron of mass '$$m$$' and a photon have same energy '$$E$$'. The ratio of de-Broglie wavelength of electron to the wavelength of photon is $$(\mathrm{c}=$$ velocity of light)

Accepted Answer

To determine the ratio of the de-Broglie wavelength of an electron to the wavelength of a photon when both have the same energy $$E$$, we need to understand their respective relations with energy and wavelength.
The de-Broglie wavelength $$\lambda_e$$ of an electron is given by:
$$ \lambda_e = \frac{h}{p} $$
where $$h$$ is Planck's constant and $$p$$ is the momentum of the electron. For an electron with non-relativistic energy, its kinetic energy $$E$$ is related to its momentum by:
$$ E = \frac{p^2}{2m} $$
Thus, we can solve for the momentum:
$$ p = \sqrt{2mE} $$
Substituting this into the de-Broglie wavelength formula, we get:
$$ \lambda_e = \frac{h}{\sqrt{2mE}} $$
Now, for a photon, its wavelength $$\lambda_{\gamma}$$ is related to its energy by:
$$ E = \frac{hc}{\lambda_{\gamma}} $$
Solving for the wavelength, we get:
$$ \lambda_{\gamma} = \frac{hc}{E} $$
We need the ratio of the de-Broglie wavelength of the electron to the wavelength of the photon:
$$ \frac{\lambda_e}{\lambda_{\gamma}} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}} $$
Simplifying the expression, we get:
$$ \frac{\lambda_e}{\lambda_{\gamma}} = \frac{1}{c} \sqrt{\frac{E}{2m}} $$
Thus, the correct answer is:
Option C: $$\frac{1}{c} \sqrt{\frac{E}{2 m}}$$

Answer

$$c \sqrt{\frac{E}{m}}$$

Answer

$$\frac{1}{c} \sqrt{\frac{2 m}{g}}$$

Answer

$$\frac{1}{c} \sqrt{\frac{E}{2 m}}$$

Answer

$$c \sqrt{\frac{m}{E}}$$

Dual Nature of Matter and Radiation

An electron of mass '$m' and a photon have same energy 'E'. The ratio of de-Broglie wavelength of electron to the wavelength of photon is (\mathrm{c}=$ velocity of light)

Choose the correct answer: