Question

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

Answer 1

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 .

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