MCQ
The de Broglie wavelength of a tennis ball of $60\, g$ moving with a velocity of $10$ meters per second is approximately
- A$10^{-16}$ meters
- B$10^{-25}$ meters
- ✓$10^{-33}$ meters
- D$10^{-31}$ meters
Mass of ball $m=60 \mathrm{g}$
Velocity $v=10 \mathrm{m} / \mathrm{s}$
Plank constant $h=6.63 \times 10^{-34} \mathrm{Js}$
We know that,
$\lambda=\frac{h}{m v}$
$\lambda=\frac{6.6 \times 10^{-34}}{60 \times 10^{-3} \times 10}$
$\lambda=1.1 \times 10^{-33}$
$\lambda=10^{-33} \mathrm{m}$
Hence, the wavelength is $10^{-33} \mathrm{m}$
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$1\,\,\,$-$\,\,\,2\,\,\,$-$\,\,\,3\,\,\,$-$\,\,\,4$
(Where. $P =$ polar, $NP =$ non-polar)