MCQ
Calculate the $\lambda$ of $CO_2$ molecule moving with a velocity $440\, m/s.$
- A$\lambda = 1.03 \times 10^{-11}$
- B$\lambda = 2.06 \times 10^{-10}$
- C$\lambda = 4.12 \times 10^{-11}$
- ✓$\lambda = 2.06 \times 10^{-11}$
✓
Answer
Correct option: D.
$\lambda = 2.06 \times 10^{-11}$
d
The wavelength is calculated from the de-Broglie's equation. $\lambda=\frac{h}{m u}=\frac{6.626 \times 10^{-34}}{\frac{44}{1000 \times 6.023 \times 10^{23}} \times 440}=2.06 \times 10^{-11}$
The wavelength is calculated from the de-Broglie's equation. $\lambda=\frac{h}{m u}=\frac{6.626 \times 10^{-34}}{\frac{44}{1000 \times 6.023 \times 10^{23}} \times 440}=2.06 \times 10^{-11}$
Thus, the wavelength of $C O_{2}$ is $2.06 \times 10^{-11} \mathrm{m}$
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