CBSE 2015, Set 3

Question

The equivalent wavelength of a moving electron has the same value as that of a photon of energy $6 	imes 10^{-17}$ J. Calculate the momentum of the electron.

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** According to the de Broglie hypothesis, the momentum of an electron is given by: $$p = \frac{h}{\lambda}$$ Where: $h = 	ext{Planck's constant} = 6.626 	imes 10^{-34} 	ext{ J s}$ $\lambda = 	ext{Wavelength of electron}$ Energy of photon: $$E = \frac{hc}{\lambda} = 6 	imes 10^{-17} 	ext{ J}$$ Where: $c = 	ext{Speed of light} = 3 	imes 10^8 	ext{ m/s}$ $\lambda = 	ext{Wavelength of photon}$ Therefore, $$\lambda = \frac{hc}{E}$$ $$\lambda = \frac{(6.626 	imes 10^{-34}) (3 	imes 10^8)}{6 	imes 10^{-17}}$$ $$\lambda = 3.313 	imes 10^{-9} 	ext{ m}$$ Substituting the value of $\lambda$ in the momentum equation: $$p = \frac{6.626 	imes 10^{-34}}{3.313 	imes 10^{-9}}$$ $$p = 2 	imes 10^{-25} 	ext{ kg m s}^{-1}$$

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Physics > Modern Physics > Dual Nature Of Radiation And Matter

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