Q. The de Broglie wavelength of an electron moving with a velocity 1.5 × 10^8 m/s is equal to that of a photon calculate the ratio of the kinetic energy of the electron to that of a Photon .
Answers
Answer:-
Broglie wavelength of an electron moving with a velocity of 1.5×10^8 m/s is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon:
The de-Broglie wavelength of an electron moving with a velocity of 1.5×10^8 m/s is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon:
By GK Planet Team Saturday, 16 December 2017
Q. The de-Broglie wavelength of an electron moving with a velocity of
1.5
×
10
8
m
/
s
1.5×108 m/s is equal to that of a photon.The ratio of kinetic energy of the electron to that of the photon
(
c
=
3
×
10
8
m
/
s
)
(c=3×108 m/s):
Solution
Matter waves are a central part of the theory of quantum mechanics, being an example of wave–particle duality. All matter can exhibit wave-like behavior. For example, a beam of electrons can be diffracted just like a beam of light or a water wave.
The de Broglie equation is an equation used to describe the wave properties of matter, specifically, the wave nature of the electron: λ = h/mv, where λ is wavelength, h is Planck's constant, m is the mass of a particle, moving at a velocity v. de Broglie suggested that particles can exhibit properties of waves.
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here, v= 1.5×10^8 m/s
lemda = h/ mv
m= h/ v lemda
kinetic energy of electron ,
= 1/2 mv^2
= 1/2 h/ v lemda × v^2
= hv/ 2× lemda
kinetic energy of photon ,
= hv= hc/ lemda
Ke/Kp = hv/2 lemda × lemda / hc
= v/2c
= 1.5×10^8/2×3×10^8
= 1/4
Ke = Kp/ 4