E of the ground state = 13.6 eV
E of the n=2 state = 3.4 eV
So the energy required is 10.2 eV or about 6.367 *10^19 J
When I set this equal to 1/2 mv^2, I get a huge number for v...like numbers *10^23. That's incredibly far from the listed answer. What am I doing wrong?
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?
Ke = 8.99 *10^9
e = 1.602 *10^-19
ħ = 1.05 * 10 ^-34
Mass of electron =...