If the deBroglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6 \times 10^{14} \mathrm{~Hz}$, then the speed of electron is equal to :
(Speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ )
Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}$
Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )
Correct Option: , 4
(4) de-Broglie wavelength,
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=10^{-3}\left(\frac{3 \times 10^{8}}{6 \times 10^{14}}\right) \quad\left[\because \lambda=\frac{\mathrm{c}}{\mathrm{v}}\right]$
$\mathrm{v}=\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{9.1 \times 10^{-31} \times 3 \times 10^{5}}$
$\mathrm{v}=1.45 \times 10^{6} \mathrm{~m} / \mathrm{s}$
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