Posted by Jenny on .
Please help me answer the question below and also show equations and steps of how you came to your answer. Thank you
1. The electron in a hydrogen tom is in the first excited state, when the electron acquires an additional 2.86eV of energy. What is the quantum number n of the state into which the electron moves?
2. A singly ionized helium atom (He+) has only one electron in orbit around the nucleus. What is the radius of the ion when it is in the second exited state?
3. Find the energy (in joules) of the photon that is emitted when the electron in a hydrogen atom undergoes a transition from the n=7 energy level to produce a line in the Paschen series.

Physics (1 of 3) 
drwls,
Those are three questions, not one.
1. The first excited state of hydrogen is the n=2 level and that has an energy of 13.6/2^2 = 3.40 eV. Adding 2.86 eV to that results in an energy of 0.54 eV
The quantum level n' of that state is given by
13.6/n'^2 = 0.54
Solve for n' and round off to the nearest integer. Orbital quantum numbers must be integers.
2. Look for the equation that tells you how the "Bohr" radius of a hydrogenic (oneelectron, Rydberg) atom depends upon the nuclear atomic number, Z (which in this case, is 2) and the principal quantum number (in this case, also 2).
http://en.wikipedia.org/wiki/Rydberg_atom
has the formula you need. 
Physics (#3 of 3) 
drwls,
H9int: For the Paschen series, the quantum number of the lower state is n = 3. The upper state of the line in question is n=7. For the photon energy, use the formula
E = 13.6 eV * (1/3^2  1/7^2)