A hydrogen atom is in state N = 4, where N = 1 is the lowest energy state. What

Question

A hydrogen atom is in state N = 4, where N = 1 is the lowest energy state. What is K+U in electron volts for this atomic hydrogen energy state?

A hydrogen atom is in state N 4, where N-1 is the lowest energy state. What is K+U in electron volts for this atomic hydrogen energy state? E4 = The hydrogen atom makes a transition to state N 1. What is K+U in electron volts for this lower atomic hydrogen energy state? E1 = What is the energy in electron volts of the photon emitted in the transition from level N-4 to N1? Ephoton = Which of the graphs below represents this transition? Select- eV eV eV

Explanation / Answer

formula EN = K+U = -13.6ev / N^2, where N can equal 1, 2, 3,4 etc.

Energy in volts of the photon emitted = (13.6eV) [ 1/(n_f)^2 - 1/(n_i)^2 ]

where n_f is final energy level and likewise, n_i is initial energy level.

a) -1 * (13.6 / n^2)

ex: E2 = -1 * (13.6 / 4^2 ) = -13.6 / 4 = -0.85

b)

. When the atom transitions to energy state N=1, K+U = (-13.6ev)/ (1)^2. This should come out to be -13.6ev / 1, which should then equal -13.6 ev.

E1 = -1 * 13.6 / 1 = -13.6

c)   

When the electon transfers from level N = 4 to N = 1, simply subtract the energies of those levels. The set up should look like: Etotal = (E at N=4) - (E at N=1)
The math for this should look something like this: -0.85 - (-13.6), which equals -0.85 + 13.6, which should then equal 12.75 ev

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