1A. Find the energy of a triply ionized beryllium atom, whose electron is in the

Question

1A. Find the energy of a triply ionized beryllium atom, whose electron is in the following state: n = 2. Answer in eV.

NOTE: Before answering check other questions similar to this one. All the other ones are incorrect. For example IF n=5 the answer is not 2.176. So use the wrong example as a means to check your answer, use the same method as you did for n=2 for n=5 and if 2.176 is the answer then you did something wrong.

1B.  In the hydrogen spectrum, the series of lines called the Brackett series results from transitions to the n = 4 energy level. What is the shortest wavelength in this series? Answer in nm.

Explanation / Answer

E = R*(Z^2)*(1/n2)

where Z=4 and R=1.097X10^7 m^-1.

E = 1.097*107*42*1/22 = 4.38*10^7ev

for each energy level in an H atom, E = -Rhc/(n^2)

R = the Rydberg constant, 1.0974 x 10^7 /m
h = Planck's constant, 6.63 x 10^-34 J sec
c = speed of light, 3.0 x 10^8 m/sec
n = the energy level you are calculating the energy of

for a transition between two energy levels:

(delta)E= -Rhc(1/(n(finish)^2) - 1/(n(start)^2) (this equation comes from subtracting the above equation for two energy levels and factoring out constants)

where n(start) and n(finish) indicate the starting and ending energy levels

if the Pfund series falls back to n = 4, then the lowest E emitted will be from n=5 back to n=4...(delta)E to any n other than n=5 will be greater than from 5 to 4, so just plug and chug:

(delta)E= -(1.0974 x 10^7 )(6.63 x 10^-34)(3.0 x 10^8)(1/42-1/52)= - 4.911*0-19J (this is for one H atom - the negative sign means energy is given off...you can obviously convert this to moles by multiplying by Avogadro's number)

we know E=hv (planck's constant x frequency)
so v = 7.40 x 10^14 Hz

and we know c=(lambda)v, so lambda = 4.054 x 10^-8 m, or 40.54 nm, which is definitely an infrared wavelengt

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