What is momentum of a microscopic particle having de Broglie's wavelength $6.0 Å$ ? ( $\mathrm{h}=6.63…
What is momentum of a microscopic particle having de Broglie's wavelength $6.0 Å$ ? ( $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$ )
- $4.6 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}$
- $1.1 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}$
- $3.18 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}$
- $6.36 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}$
Solution
According to the de Broglie's equation,
$\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\
\therefore \quad \mathrm{p} & =\frac{\mathrm{h}}{\lambda}=\frac{6.63 \times 10^{-34} \mathrm{Js}}{6.0 \times 10^{-10} \mathrm{~m}}=1.1 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$
Asked in: MHT CET 2024 (02 May Shift 2)
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