KCET · Chemistry · Structure of Atom
A body of mass \(10 \mathrm{mg}\) is moving with a velocity of \(100 \mathrm{~ms}^{-1}\). The wavelength of de-Broglie wave associated with it would be ( \(h=6.63 \times 10^{-34} \mathrm{Js}\) )
- A \(6.63 \times 10^{-35} \mathrm{~m}\)
- B \(6.63 \times 10^{-34} \mathrm{~m}\)
- C \(6.63 \times 10^{-31} \mathrm{~m}\)
- D \(6.63 \times 10^{-37} \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(6.63 \times 10^{-31} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(m=10 \mathrm{mg}=10 \times 10^{-6} \mathrm{~kg}\)
\[
\begin{aligned}
v &=100 \mathrm{~ms}^{-1} \\
\lambda &=\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{10 \times 10^{-6} \times 100} \\
&=6.63 \times 10^{-31} \mathrm{~m}
\end{aligned}
\]
\[
\begin{aligned}
v &=100 \mathrm{~ms}^{-1} \\
\lambda &=\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{10 \times 10^{-6} \times 100} \\
&=6.63 \times 10^{-31} \mathrm{~m}
\end{aligned}
\]
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