is potential energy and kinetic energy real or is it theoretical and is used to explain how thing work? if it is real than where
1 answer:
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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) .
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Answer:
v = 54.2 m / s
Explanation:
Let's use energy conservation for this problem.
Starting point Higher
Em₀ = U = m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 150)
v = 54.2 m / s