Problem Set (2) By Sayantan Bhattacharya

The entropy Sof a system of N spins, which may align either in the upward or in the downward direction, is given by S--kNIpln p+(1 - p)ln(1-p)] Herek is the Boltzmann constant. The probability of alignment in the upward direction isp. Ihe value of p, at which the entropy is maximum, is Give your answer up-to one decimal place)[GATE 2016]

. So, we need to find p value at maximum of S os S to be maximum0 op SkNIp ln p +(1-p)ln(1 -p)] os op 1- P 1- p -> p = 0.5

GATE2016 Consider a system having three energy levels with energies 0, 2E and 3e ,with respective degeneracy of 2, 2 and 3. Four bosons of spin zero have to be accommodated in these levels such that the total energy of the system is 10 The number of ways in which it can be done is

. The system have energy 10e,it can be possible if out of four boson two boson are in energy level 2 and two boson are in energy level 3 , the possible ways of arranging is basically the number of microstates: n gi-1 2+2-1 2+3-1 22-1 23-1 18

GATE2016 Atwo-level system has energies zero and E. The level with zero energy is nondegenerate, while the level with energy Eistriply degenerate. The mean energy of a classical particle in this system at atemperature Tis: Soln: . KE) - 8ie E,IkT - E IkT 0/kT E IkT 3eEIKT

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Sayantan Bhattacharya

"Educating India For a Better Tomorrow"
||Ph.D student,U mass Lowell,Massachusetts,USA||
M.Sc,University Of Hyderabad,2018||
B.Sc,B.H.U,2016

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Vandana Prajapati

a year ago

math ki video

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Sanjeev Mishra

a year ago

bht jld hi provide ki jayengi
abhi English Grammar ka course preapre ho rha jiske videos aapko parson se milne lagenge

Vandana Prajapati

a year ago

ok sir

Vandana Prajapati

a year ago

thanks