The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)
- 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 .
This can be written as .....(1) where λ is known as de Broglie wavelength and p is momentum , h = Plank’s constant .
As we know that mass of proton is greater than electron and photon .
(a)For photon , the momentum is given by ...(2) where c is the speed is the speed of light .
Putting E = 3eV in equation (2) , we get
Putting this value of p in equation (1) , we get
(b) As we know that kinetic energy is given by
...(3)
Where mass of electron is .
Putting K.E = 3eV in equation (3) , we get
Putting this value of p in equation (1) , we get
(c) Putting and K.E = 3eV in equation (3) , we get
Putting this value of p in equation (1) , we get
(d) Putting E = 0.3eV in equation (2) , we get
Putting this value of p in equation (1) , we get
(e) Putting in equation (1) , we get
On comparing the wavelength order should be (d) > (a) = (e) > (b) > (c) .
Learn more about de brogile here :
#SPJ4