The wavelength ' λ ' of a photon and the deBroglie wavelength of an… | MHT CET 2025 (26 Apr, Shift 1)

Q35MHT CET 2025 · 26 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

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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Answer: (c)

Same $\lambda$ means same $p = h/\lambda$ .
Electron: $K = \dfrac{h^2}{2m\lambda^2}$ . Photon: $E = \dfrac{hc}{\lambda}$ .
$\dfrac{K}{E} = \dfrac{h}{2\lambda mc}$
So the answer is (c).

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