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If the deBroglie wavelength of an electron is equal to $10^{-}$ 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 $\left.=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}$
Mass of electron $\left.=9.1 \times 10^{-31} \mathrm{~kg}\right)$
Options:
(Speed of light $\left.=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}$
Mass of electron $\left.=9.1 \times 10^{-31} \mathrm{~kg}\right)$
Solution:
2416 Upvotes
Verified Answer
The correct answer is:
$1.45 \times 10^{6} \mathrm{~m} / \mathrm{s}$
de-Broglie wavelength,
$$
\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=10^{-3}\left(\frac{3 \times 10^{8}}{6 \times 10^{14}}\right) \quad\left[\because \lambda=\frac{\mathrm{c}}{\mathrm{v}}\right]
$$
$$
\begin{array}{l}
\mathrm{v}=\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{9.1 \times 10^{-31} \times 3 \times 10^{5}} \\
\mathrm{v}=1.45 \times 10^{6} \mathrm{~m} / \mathrm{s}
\end{array}
$$
$$
\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=10^{-3}\left(\frac{3 \times 10^{8}}{6 \times 10^{14}}\right) \quad\left[\because \lambda=\frac{\mathrm{c}}{\mathrm{v}}\right]
$$
$$
\begin{array}{l}
\mathrm{v}=\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{9.1 \times 10^{-31} \times 3 \times 10^{5}} \\
\mathrm{v}=1.45 \times 10^{6} \mathrm{~m} / \mathrm{s}
\end{array}
$$
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