A proton, an electron, and a helium nucleus all move at speed v . Rank their de Broglie wavelengths from largest to smallest.

Physics

1 answer:

Varvara68 [4.7K]1 year ago

8 0

Ranking of de Broglie wavelengths from largest to smallest is electron > proton > helium

De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship

De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ .....(1) , where λ is known as de Broglie wavelength and m is mass , v is velocity , h = Plank’s constant.

From equation (1) wavelength and mass has an inverse relation .

Mass of helium is 4 times the mass of the proton and proton has a greater mass than electron.

According to equation (1) , less the mass higher will be the wavelength

Hence electron having less mass have higher wavelength and then proton and then helium having large mass will have less wavelength .

Thus, order should be electron > proton > helium .

Learn about de brogile wavelength more here :

brainly.com/question/16595523

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Answer:

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