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(C) The momentum of a photon is given by $p = \frac{h}{\lambda}$.Given wavelength $\lambda = 5200 \, \mathring{A} = 5200 	imes 10^{-10} \, m$.The mom

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$1400$

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(C) The momentum of a photon is given by $p = \frac{h}{\lambda}$.Given wavelength $\lambda = 5200 \, \mathring{A} = 5200 	imes 10^{-10} \, m$.The momentum of the electron is $p = m_e v$.Equating the two momenta: $m_e v = \frac{h}{\lambda}$.Therefore, $v = \frac{h}{m_e \lambda}$.Substituting the values: $v = \frac{6.63 	imes 10^{-34}}{(9.1 	imes 10^{-31}) 	imes (5200 	imes 10^{-10})}$.$v = \frac{6.63 	imes 10^{-34}}{47.32 	imes 10^{-22}} \approx 0.1401 	imes 10^{-12} 	imes 10^{22} \approx 1401 \, m/s$.Rounding to the nearest option, the velocity is $1400 \, m/s$.

What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength $5200 \, \mathring{A}$? (Given: $h = 6.63 \times 10^{-34} \, J \cdot s$, $m_e = 9.1 \times 10^{-31} \, kg$)

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