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A hypothetica atom has energy levels uniformaly separated by 1.3 eV. At a temperature 2500k, what is the ratio of number of atoms in 15th excited state to the number in 13 th excited state.

Answer» It is known that in excited state x, no. of atoms, `N_x=N_0e^((E_x-E_0)//kT)`
`:. (N_(15))/(N_(13))=e^(-(E_(15)-E_(13))//kT)`
`=exp.[-(2xx1.3eV)/(kT)]`
`=exp.[-(2.6xx1.6xx10^(-19))/(1.38xx10^(-23)xx2500)]`
`(N_(15))/(N_(13))=exp.[-12]=1/(e^(12))=6.15xx10^-6`


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