Question

A system consist of N no of particles ,N>>1.each particle can have only one of the two energies E1or E2+ε(ε>0).if the system is in equllibrium at a temperature T,the avg no. of particles with energy E1 is.....

Answer 1

the requested probability, is given by the subsequent formula:
\[\huge p = \frac{{{e^{ - {E_1}/kT}}}}{{{e^{ - {E_1}/kT}} + {e^{ - \left( {{E_1} + \varepsilon } \right)/kT}}}} = ...?\]
where \(E_1\) and \(E_1+\epsilon\), are the energies of such quantum states

Answer 2

it is the probability that the system lies in quantum state with energy \(E_1\)

Answer 3

with such probability, it is easy to compute the requested average number of particles.
Furthermore, \(k\) is the \(Boltzmann\) constant

Answer 4

sry bt i didnt get this .....wht is d ans & will u tell plz me wht does question represents.......

Answer 5

options are like....1.N/2 2.N/e^(ε/KT+1) 3.N/e^(-ε/KT+1) 4.Ne^(-ε/KT)

Answer 6

the requested number is given by the subsequent product:
\[\Large \begin{gathered}
Np = \frac{{N{e^{ - {E_1}/kT}}}}{{{e^{ - {E_1}/kT}} + {e^{ - \left( {{E_1} + \varepsilon } \right)/kT}}}} = \hfill \\
\hfill \\
= N{e^{ - {E_1}/kT}}\frac{1}{{{e^{ - {E_1}/kT}}\left( {1 + {e^{ - \varepsilon /kT}}} \right)}} = ...? \hfill \\
\end{gathered} \]
please continue
@pratima99