Question
Question: The de broglie wavelength of 1 mg grain of sand blown by a 20ms _1 wind is : A.. \[3.3*{{10}^{-21}...
The de broglie wavelength of 1 mg grain of sand blown by a 20ms _1 wind is :
A.. 3.3∗10−21m
B. 3.3∗10−21m
C. 3.3∗10−49m
D. 3.3∗10−42m
Solution
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 ∼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∗10−6kg
v=20m/s
h=6.626∗1034J.s
Now,
λ=20∗10−66.626∗10−34
λ=3.313∗10−29m
The de Broglie wavelength of 1 mg grain of sand blown at 20m/s wind is 3.3∗10−29m
So, The correct option is A. 3.3∗10−29m
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.