A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W
1 answer:
Answer:
Explanation:
a ) Slit separation d = .1 x 10⁻³ m
Screen distance D = 4 m
wave length of light λ = 650 x 10⁻⁹ m
Width of central fringe = λ D / d
=
= 26 mm
b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm
c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to
λ / d
=
= 6.5 x 10⁻³ radian.
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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 .....(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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