Question 264: Dual Nature of Matter and Radiation

Question

The wavelength ' $\lambda$ ' of a photon and the deBroglie wavelength of an electron have same value. the ratio of kinetic energy of the electron to the energy of a photon is
( $\mathrm{m}=$ mass of electron, $\mathrm{c}=$ velocity of light, $\mathrm{h}=$ Planck's constant)

Accepted Answer

A photon and an electron both have the same wavelength :
$$ \lambda_{	ext {photon }}=\lambda_{	ext {wavelength }}=\lambda $$
Photon Energy : $\mathrm{E}_{	ext {photon }}=\frac{\mathrm{hc}}{\lambda}$
Electron's de-Broglie wavelength,
$$ \lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m K . E .}} \Rightarrow 	ext { K.E. }=\frac{h^2}{2 m \lambda^2} $$
Ratio of electron $\mathrm{K} \cdot \mathrm{E}$ to photon energy,
$$ \begin{aligned} & \frac{	ext { K.E. }}{\mathrm{E}_{	ext {photon }}}=\frac{\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda^2}}{\frac{\mathrm{hc}}{\lambda}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda} \cdot \frac{1}{\mathrm{c}} \ 	herefore \quad & \frac{	ext { K.E. }}{\mathrm{E}_{	ext {photon }}}=\frac{\mathrm{h}}{2 \lambda \mathrm{mc}} \end{aligned} $$

Answer

$\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$

Answer

$\frac{\lambda \mathrm{mc}}{\mathrm{h}}$

Answer

$\frac{\mathrm{h}}{2 \lambda \mathrm{mc}}$

Answer

$\frac{\mathrm{h}}{\lambda \mathrm{mc}}$

Dual Nature of Matter and Radiation

The wavelength ' \lambda ' of a photon and the deBroglie wavelength of an electron have same value. the ratio of kinetic energy of the electron to the energy of a photon is ( \mathrm{m}= mass of electron, \mathrm{c}= velocity of light, \mathrm{h}= Planck's constant)

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