If the de-Broglie wavelength of an electron is equal to 10-3 times the wavelength of a photon of frequency 6×1014 Hz, then the speed of electron is equal toSpeed of light =3×108 ms-1Planck's constant =6.63×10-34 JsMass of electron =9.1×10-31 kg

λphoton=cν=3×1086×1014=0.5×10-6 mSince, λelectron=λphoton103=0.5×10-9 m⇒ hmv=λelectron ⇒ v=hmλelectron=6.63×10-349.1×10-310.5×10-9=1.46×106 ms-1

If the de-Broglie wavelength of an electron is equal to 10-3 times the wavelength of a photon of frequency 6×1014 Hz, then the speed of electron is equal toSpeed of light =3×108 ms-1Planck's constant =6.63×10-34 JsMass of electron =9.1×10-31 kg

λphoton=cν=3×1086×1014=0.5×10-6 mSince, λelectron=λphoton103=0.5×10-9 m⇒ hmv=λelectron ⇒ v=hmλelectron=6.63×10-349.1×10-310.5×10-9=1.46×106 ms-1

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If the de-Broglie wavelength of an electron is equal to $10^{- 3}$ times the wavelength of a photon of frequency $6 \times 10^{14} Hz$ , then the speed of electron is equal to
Speed of light $= 3 \times 10^{8} {ms}^{- 1}$
Planck's constant $= 6.63 \times 10^{- 34} Js$
Mass of electron $= 9.1 \times 10^{- 31} kg$

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$1.7 \times 10^{6} {ms}^{- 1}$

$1.46 \times 10^{6} {ms}^{- 1}$

$1.8 \times 10^{6} {ms}^{- 1}$

$1.1 \times 10^{6} {ms}^{- 1}$

answer is B.

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Detailed Solution

$λ_{photon} = \frac{c}{ν} = \frac{3 \times 10^{8}}{6 \times 10^{14}} = 0.5 \times 10^{- 6} m$

Since, $λ_{e l e c t r o n} = \frac{λ_{photon}}{10^{3}} = 0.5 \times 10^{- 9} m$

$\Rightarrow \frac{h}{m v} = λ_{e l e c t r o n} \Rightarrow v = \frac{h}{m λ_{e l e c t r o n}} = \frac{6.63 \times 10^{- 34}}{(9.1 \times 10^{- 31}) (0.5 \times 10^{- 9})} = 1.46 \times 10^{6} {ms}^{- 1}$

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If the de-Broglie wavelength of an electron is equal to 10-3 times the wavelength of a photon of frequency 6×1014 Hz, then the speed of electron is equal toSpeed of light =3×108 ms-1Planck's constant =6.63×10-34 JsMass of electron =9.1×10-31 kg

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