Generally useful constants (MKS):amu = 1.6605 H10^−27L kgNA = 6.02221 H10^23Lêmolek = 1.38066 H10^−23L J ê Kme = 9.1094 H10^−31L kgh = 6.62608 H10^−34L Jsq = 1.6022 H10^−19L Cc = 2.9979 H10^8L m ê s1.6605 × 10−27kg6.02221 × 1023mole1.38066 × 10−23JK9.1094 × 10−31kg6.62608× 10−34Js1.6022 × 10−19C2.9979 × 108msProblem 2.1 (Engel)Assume that a system has a very large number of energy levels given by the formula e = e0 l2with e0= 2.34 μ 10-22J,where l takes on the integer values 1, 2, 3, ... . Assume also that the degeneracy of a level is given by gi= 2 l. Calcu-late the population ratios n5ên1and n10ê n1for T = 100 K and 650 K, respectively.SolutionStrategy. Population ratios are obtained by assuming a Boltzmann distribution (which holds only if the system is atthermal equilibrium).niÅÅÅÅÅÅÅÅn1=giÅÅÅÅÅÅÅÅg1 ‰-Hei-e1LêkTPrediction. Populations should fall as state energy increases. In other words, n1> n5> n10 should hold at bothtemperatures. However, the populations should fall less at higher temperature.Execution. First, calculate ei-e1e0 = 2.34 ∗ 10^−22 Je51 = e0 ∗ H5^2− 1^2Le101 = e0 ∗ H10^2 − 1^2L2.34 × 10−22J5.616 × 10−21J2.3166 × 10−20JNext, calculate kTT = 8100 K, 650 K<kT = kT8100 K, 650 K<81.38066× 10−21J, 8.97429 × 10−21J<Next, calculate the ratio of state degeneracies, giêg1g51 = H2 ∗ 5LêH2 ∗ 1Lg101 = H2 ∗ 10LêH2 ∗ 1L5— General::spell1 : Possible spelling error: new symbolname "g101" is similar to existing symbol "e101".More…10Ignore warning about similarity in variable names. Calculate n5ên1 at both temperatures2P2.1.nbg51  ^H−e51 ê kTL80.0855904, 2.6742<Calculate n10ê n1 at both temperaturesg101  ^H−e101 ê kTL85.16419× 10−7, 0.7567<CommentPart of my prediction was flawed. It is possible for excited states to have higher populations than lower energy states,but only if the excited states have higher degeneracies. Thus, at 650 K, n5 > n1.The other parts of my prediction worked out as