If the de-Broglie wavelength of an electron is equal to 10−4 times the wavelength of a photon of frequency 6×1016Hz, then the speed of electron is equal to (speed of light = 3×108m/s ; planck’s constant = 6.63×10−34 Js, mass of electron = 9.1×10−31kg )

λp=cV=3×1086×1016=12×10−8⇒λe=10−4λp=12×10−12λ=h/pV=hλm=6.63×10−3412×10−12×9.1×10−31=1.45×109 m/s

If the de-Broglie wavelength of an electron is equal to 10−4 times the wavelength of a photon of frequency 6×1016Hz, then the speed of electron is equal to (speed of light = 3×108m/s ; planck’s constant = 6.63×10−34 Js, mass of electron = 9.1×10−31kg )

λp=cV=3×1086×1016=12×10−8⇒λe=10−4λp=12×10−12λ=h/pV=hλm=6.63×10−3412×10−12×9.1×10−31=1.45×109 m/s

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If the de-Broglie wavelength of an electron is equal to $10^{- 4}$ times the wavelength of a photon of frequency $6 \times 10^{16} H z,$ then the speed of electron is equal to
(speed of light = $3 \times 10^{8} m / s$ ; planck’s constant = $6.63 \times 10^{- 34}$ Js, mass of electron = $9.1 \times 10^{- 31} k g$ )

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$1.45 \times 10^{6} m / s$

$1.45 \times 10^{9} m / s$

$1.7 \times 10^{7} m / s$

$1.1 \times 10^{6} m / s$

answer is B.

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Detailed Solution

$\begin{array}{l} λ_{p} = \frac{c}{V} = \frac{3 \times 10^{8}}{6 \times 10^{16}} = \frac{1}{2} \times 10^{- 8} \\ \Rightarrow λ_{e} = 10^{- 4} λ_{p} = \frac{1}{2} \times 10^{- 12} \\ λ = h / p \\ V = \frac{h}{λ m} = \frac{6.63 \times 10^{- 34}}{\frac{1}{2} \times 10^{- 12} \times 9.1 \times 10^{- 31}} = 1.45 \times 10^{9} m / s \end{array}$