1. The problem statement, all variables and given/known data Two hydrogen atoms, both initially in the ground state, undergo a head-on collision. If both atoms are to be excited to the n = 2 level in this collision, what is the minimum speed each atom can have before the collision? Answer in units of m/s 1 mol of hydrogen is 1.008g 2. Relevant equations En = K + Uelec For a hydrogen atom: En= 13.6/N^2 3. The attempt at a solution Okay, so I know that there is a change in energy that excites the atoms to N=2. This change in energy must be from kinetic energy. So, Two atoms collide into each other. And after the collision they are raised to N=2. A. 2(Ei+K)=2(Ef) ;note, i used 2 because it is two atoms colliding The twos cancel? B. Ei+K=Ef -13.6/(1*1)+1/2(mh)v^2=-13.6/(2*2) ;mh is mass of hydrogen -13.6+1/2(mh)V^2=-3.4 C. 10.2=1/2(mh)v^2 ;add 13.6 to both sides Am I correct up until here? (Part C) To find mass of hydrogen D. 1.008/6.02*10^23=1.67*10^-24 E. 10.2=1/2(1.67*10^-24)*(v^2) 20.4=(1.67*10^-24)*(V^2) 1.22155 E 25=V^2 Where did I mess up?