En=n2−13.6=−0.54 n=5n = 5n=5no of transitionn(n−1)2=5(5−1)2=10\frac{n(n-1)}{2} = \frac{5(5-1)}{2} = 102n(n−1)=25(5−1)=10 This shows the calculation of the energy level of hydrogen at n=5n=5n=5, and the number of possible electronic transitions (101010) from that level.