Question

The de Broglie wavelength of an electron moving with a velocity 1.5 × 10⁸ m/s is equal to that of a photon. The ratio of the energy of the photon to the kinetic energy of the electron is ______.

Answer 1

Correct Answer: 4

Explanation:
Given:Velocity of the electron , v = 1.5 × 10⁸ m/sVelocity of the photon , c = 3 × 10⁸ m/sde Broglie wavelength of the electron is equal to that of a photon, i.e.${\lambda }_e={\lambda }_p$We have to find the ratio of the energy of the photon and the kinetic energy of the electron.From the given values of v and c, we have$\frac{c}{v}=\frac{3\times {10}^{8}m/s}{1.5\times {10}^{8}m/s}=2$...(i)de Broglie wavelength of electron is${\lambda }_e=\frac{h}{m_{e}v}$$\Rightarrow m_{e}v=\frac{h}{{\lambda }_e}$ ..(ii)Kinetic energy of the electron is$k=\frac{1}{2}{m}_e{v}^2$$=\frac{1}{2}\left(m_{e}v\right)v=\frac{hv}{2{\lambda }_e}$ ..(iii) [from eq. (ii)]Let $\lambda p$ be the wavelength of the photon, then energy of the photon is$E=h{v}_p=\frac{hc}{{\lambda }_p}$ ...(iv)where, $v_p=\frac{c}{{\lambda }_p}$From eqs. (iii) and (iv), we have $\frac{E}{K}=\frac{hc/{\lambda }_p}{hv/2{\lambda }_e}$$=\left(\frac{c}{v}\right)\left(\frac{2{\lambda }_e}{{\lambda }_p}\right)$Using eq. (i), we get$\frac{E}{K}=2\times 2=4\text{ [given, }{\lambda }_e={\lambda }_p]$Hence, the correct answer is 4.00.