An energy of 13.6 eV is needed to ionize an electron from the ground state of a

Question

An energy of 13.6 eV is needed to ionize an electron from the ground state of a hydrogen atom. Selecting the longest wavelength that will work from the those given below, what wavelength is needed if a photon accomplishes this task? What value of wavelength is associated with the Lyman series for n = 2? An electron in a hydrogen atom makes a transition from the n = 3 to the n = 1 energy state. Determine the wavelength of the emitted photon (in nm). A Li^2+ ion undergoes a transition from the n - 4 to the n - 3 state. The energy of the emitted photon is

Explanation / Answer

I am allowed to answer only 1 question at a time but i am answering 2

7.
E = 13.6 eV
= 13.6 * 1.6*10^-19 J
= 2.176*10^-18 J

E = h*c/lambda
2.176*10^-18 = (6.626*10^-34 * 3*10^8)/ lambda
lambda = 9.1*10^-8 m= 91 nm
AnsweR: d 90 nm

9.
1st calculate the enrgy difference between n=1 and n=3
E = 13.6*[1/1^2 - 1/3^2]
=12.09 eV
=12.09*1.6*10^-19 J
= 1.93*10^-18 J

Now use:
E = h*c/lambda
1.93*10^-18 = (6.626*10^-34 * 3*10^8)/ lambda
lambda = 1.03*10^-7 m= 103 nm
AnsweR: d. 103 nm

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