Q19 of 43 Page 407

What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

We have a relation to find kinetic energy of a gas molecule at any Temperature which is



Where EK is the kinetic energy of a particle at temperature T and K is boltzmann constant


K = 1.38 × 10-23 JK-1


And we know Kinetic energy of a particle is given by the relation



Where vrms is the root meansquare speed of the particle and m is the mass of the particle


So equating both equations we have



On solving we get the root meansquare speed of the particle as



Where Vrms is the root mean square speed of the particle of mass m at temperature T and K is boltzmann constant


K = 1.38 × 10-23 JK-1


Here the temperature is


T = 300K


We are given the particle is a nitrogen molecule (N2) mass of the Nitrogen atom is


mN = 14.0076 u


so mass of N2 molecule is


m = 2 × 14.0076 u = 28.0152 u


converting it to Kg


we know 1 u = 1.66 × 10-27 Kg


so mass of Nitrogen molecule is


m = 28.0152 × 1.66 × 10-27Kg = 46.50 × 10-27Kg


= 4.650 × 10-26Kg


We know de-Broglie wavelength of a particle is given by relation



where

More from this chapter

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17

Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.

 18

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

 20

(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter.

Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 1011 C kg–1.


(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

 21

(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s–1 is subject to a magnetic field of 1.30 × 10–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011C kg–1.

(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?


[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasize the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]