The de Broglie wavelength () equation is
Where, h = 6.63 x 10-34 Js is plank's constant and
p is momentum of the particle
(a) Given,
mass of electron, me = 9.1 x 10-31 kg
velocity of electron = 15 km/s = 15000 m/s
So, momentum of electron = 15000 x 9.1 x 10-31 = 1.365 x 10-26 kg.m/s
Therefore, de Brolie wavelength of electron is,
(b) Given, the kinetic energy (E) of electron = 1 eV = 1.602 x 10-19 J
Momentum of the electron can be calculated using this equation :
or,
So, de Broglie wavelength of electron is,
(c)
Given,
mass of proton, mp = 1.67 x 10-27 kg
velocity of electron = 15 km/s = 15000 m/s
So, momentum of electron = 15000 x 1.67 x 10-27 = 2.5050 x 10-23 kg.m/s
Therefore, de Brolie wavelength of electron is,
(Angstrom)
(d)
Given, the kinetic energy (E) of proton = 1 eV = 1.602 x 10-19 J
Momentum of the proton can be calculated using this equation :
or,
So, de Broglie wavelength of proton is,
(e)
Given,
mass of elephant, melephant = 10000 kg
velocity of elephant = 15 km/hr = 4.167 m/s
So, momentum of elephant = 4.167 x 10000= 4.167 x 104 kg.m/s
Therefore, de Brolie wavelength of elephant is,
(f)
Given, the kinetic energy (E) of elephant = 1 eV = 1.602 x 10-19 J
Momentum of the elephant can be calculated using this equation :
or,
So, de Broglie wavelength of elephant in this case is,
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33.3 What is the de Broglie wavelength of: (a) an electron (melectron electron 9.1 x 10-31...
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