A proton, an electron, and a helium nucleus all move at speed v . Rank their de Broglie wavelengths from largest to smallest.
1 answer:
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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Answer:
Time period for Simple pendulum,
Explanation:
The Simple Pendulum
Consider a small bob of mass is tied to extensible string of length
that is fixed to rigid support. The bob is oscillating in the plane about verticle.
Let is the angle made by string with vertical during oscillation.
Vertical component of the force on bob,
Negative sign shows that its opposing the motion of bob.
Taking as very small angle then,
Let is the displacement made by bob from its mean position ,
then,
so, ........(1)
Since, pendulum is in hormonic motion,
as we know,
where is the constant and
.........(2)
From equation (1) and (2)
Since,