Question

33.3 What is the de Broglie wavelength of: (a) an electron (melectron electron 9.1 x 10-31 kg) travelling at 15 kms-1? (b) an electron with a kinetic energy of 1 eV? (c) a proton (mproton -1.67x 10-27 kg travelling at 15kms-1? (d) a proton with a kinetic energy of 1 eV? (e) an elephant (melephant 10 tonnes) travelling at 15 km h-1? (D an elephant with a kinetic energy of 1 evi

Answer 1

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, m_e = 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, m_p = 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, m_elephant = 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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