Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1114: 73

$78700K$

We know that $E_n=\frac{-13.6}{n^2}eV$ $\implies E_1=-13.6eV$ and $E_2=\frac{-13.6}{4}=-3.4eV$ Now $\Delta E=E_2-E_1$ $\implies \Delta E=(-3.4)-(-13.6)eV=10.2eV$ $\Delta E=(10.2)(1.6\times 10^{-19}J)$ We can find the required temperature as $T=\frac{\Delta E}{(\frac{3}{2})K}$ We plug in the known values to obtain: $T=\frac{32.64}{4.14}\times 10^4$ $T=78700K$