Question

The de broglie wavelength of 1 mg grain of sand blown by a 20ms _1 wind is :
A.. \[3.3*{{10}^{-21}}m\]
B. \[3.3*{{10}^{-21}}m\]
C. \[3.3*{{10}^{-49}}m\]
D. \[3.3*{{10}^{-42}}m\]

Answer 1

Hint: All particles can show wave-like properties. The de Broglie wavelength of a particle indicates the length scale at which wave-like properties are important for that particle.

Step by step answer: A typical electron in a metal has a de Broglie wavelength is of order \[\sim 10nm\]. Therefore, we see quantum-mechanical effects in the properties of a metal when the width of the sample is around that value.
SI unit metre, \[m\]
Expressed in SI base units \[M\]
Other commonly-used unit(s) \[nm\]
De Broglie wavelength λ=mvh​
\[m=1mg=1*{{10}^{-6}}kg\]
\[v=20m/s\]
\[h=6.626*{{10}^{34}}J.s\]
Now,
\[\lambda =20*{{10}^{-66.626}}*{{10}^{-34}}\]
\[\lambda =3.313*{{10}^{-29}}m\]
The de Broglie wavelength of 1 mg grain of sand blown at 20m/s wind is \[3.3*{{10}^{-29}}m\]

So, The correct option is A. \[3.3*{{10}^{-29}}m\]

Additional Information: On the off chance that a molecule is altogether bigger than its own de Broglie frequency, or on the off chance that it is connecting with different items on a scale essentially bigger than its de Broglie frequency, at that point its wave-like properties are not perceptible. Quantize: Limit the possible values of (a magnitude or quantity) to a discrete set of values by quantum mechanical rules.

Note: Planck's examination of the emanation spectra of hot articles and the ensuing investigations into the photoelectric impact had demonstrated that light was equipped for acting both as a wave and as a molecule. It appeared to be sensible to contemplate whether electrons could likewise have a double wave-molecule nature. In 1924, French researcher Louis de Broglie (1892–1987) inferred a condition that portrayed the wave idea of any molecule.