The average kinetic energy of an atom in a monatomic ideal gas is given by KE=(3/2)kT,where...
The average kinetic energy of an atom in a monatomic ideal gas is given by KE=(3/2)kT,where k = 1.38
Homework Answers
average KINETIC ENERGY= 3/2 *K*T = 3/2*1.38*10^-23*292=6.044*10^-21J
6.044*10^-21J = hc/lamda
lamda = 6.626*10^-34*3*10^8/(6.044*10^-21)
lamda =3.28*10^-5m
K.E = (1/2)m*v^2 = (3/2)kT
or, (1/2)(6.65*10^-27)(v^2) = (3/2)(1.38*10^-23)(292)
or, v = 1348.28 m/s
now, de Broglie wavelength ,L = h/mv = (6.63*10^-34)/{(6.65*10^-27)*(1348.28)} = 7.395*10^-11 m = 0.07395 nm
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