The de Broglie wavelength of a particle of mass 1 mg moving with a velocity of $10 \mathrm{~ms}^{-1}$ is…

The de Broglie wavelength of a particle of mass 1 mg moving with a velocity of $10 \mathrm{~ms}^{-1}$ is ($h=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}$)
  1. $6.63 \times 10^{-29} \mathrm{~m}$
  2. $6.63 \times 10^{-31} \mathrm{~m}$
  3. $6.63 \times 10^{-34} \mathrm{~m}$
  4. $6.63 \times 10^{-22} \mathrm{~m}$

Solution

According to de Broglie wavelength $(\lambda)=\frac{h}{p}$ or, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\left[\right.$ Where $\mathrm{h}=$ Plank's constant $=6.63 \times 10^{-34} \mathrm{~J}$ $\mathrm{s}=$ momentum $\mathrm{m}=$ mass of moving particle $\mathrm{v}=$ velocity $]$ $\left[1 \mathrm{~J}=\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-2} 1 \mathrm{mg}=10^{-3} \mathrm{~g} 1 \mathrm{~kg}=10^3 \mathrm{~g}\right]$ $\lambda=\frac{6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}}{1 \mathrm{mg} \times 10 \mathrm{~ms}^{-1}}$ $\begin{aligned} & \lambda=\frac{6.63 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2-\mathrm{s}^{-2} \mathrm{~s}}{10^{-3} \mathrm{mg} \times 10 \mathrm{~ms}^{-1}} \\ & \lambda=\frac{6.63 \times 10^{-34} \times 10^3 \mathrm{gm}^2 \mathrm{~s}^{-1}}{10^{-2} \mathrm{~g} \mathrm{~ms}^{-1}}\end{aligned}$ $\begin{aligned} & \lambda=6.63 \times 10^{-34} \times 10^3 \times 10^2 \mathrm{~m} \\ & \lambda=6.63 \times 10^{-29} \mathrm{~m}\end{aligned}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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