Question

Consider an element that reaches its first excited state by absorption of 314.9 nm light. a) Determine the energy difference (kJ/mol) between the ground state and the first excited state. Delta E = b) If the degeneracies of the two states for the element are g*/g_0 = 2, determine N*/N_0 at 2030 K. N*/N_0 = c) By what percentage does N*/N_0 change if the temperature is raised by 20 K? d) What is N*/N_0 at 5.00 x 10^3 K? N*/N_0 =

Answer 1

a) ΔE = N_AhC/λ, where N_A=Avagadro's no, h=Planck's constant, C=Velocity of light, λ = Wavelength of absorption.

         = (6.023*10²³)*(6.625*10^-34)*(3*10⁸)/(314.9*10^-9)

           = 38*10⁴ J/mol.

    ΔE = 380 KJ/mol.

b) If the temperature of the element during excitation (T) = 2030 K

N^*/N₀ = (g^*/g₀)*e^-ΔE/RT

= 2*e^{-380/(8.314*10^-3*2030)}

= 2*1.666*10^-10

(N^*/N₀)_{T=2030 K} = 3.33*10)^-10.

c) If the temperature is raised by 20 K, i.e. T = 2050 K

N^*/N₀ = (g^*/g₀)*e^-ΔE/RT

             = 2*e^{-380/(8.314*10^-3*2050)}

                = 2*2.075*10^-10

= 4.15*10^-10.

Therefore, the difference between N^*/N₀ values at T = 2030 and 2050 is (4.15 - 3.33)*10^-10, i.e. 0.82*10^-10.

The % increment in the N^*/N₀ value when the temperature is raised by 20 K = (0.82/3.33)*100, i.e. ~ 25%

d) If T = 5*10³ K

N^*/N₀ = (g^*/g₀)*e^-ΔE/RT

= 2*e^{-380/(8.314*10^-3*5*10^3)}

= 2*1.07*10^-4

(N^*/N₀)_{T=5000 K} = 2.14*10^-4.