(c28p47) The Lyman series for a (new!?) one-electron atom is observed in a dista
ID: 1305500 • Letter: #
Question
(c28p47) The Lyman series for a (new!?) one-electron atom is observed in a distant galaxy. The wavelengths of the first four lines and the short-wavelength limit of this Lyman series are given by the energy level diagram in Figure P28.47. Based on this information, calculate (a) the energies of the ground state and first four excited states for this oneelectron atom.
A) What is the energy of the ground state?
B) What is the energy of the first excited state?
C) What is the energy of the second excited state?
D) What is the energy of the third excited state?
E) What is the energy of the fourth excited state?
F) Calculate the longest-wavelength (alpha) lines in the Balmer series for this atom.
G) Calculate the short-wavelength series limit in the Balmer series for this atom.
Explanation / Answer
energy E =hC/?
where h is Plank's constant =6.625*10^-34 J s
C is speed of the light =3*10^8m/s
the product of h*C =(6.625*10^-34 J s)(3*10^8m/s) (1A0/ 10^-10 m)(1 eV/1.6*10^-19 J)
1 eV =1.6*10^-19 J
1 A0=10^-10 m
hc = 12400 eV-A0
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E? - E1 = hc/? where ? =1520 A0 and E? =0
E1 = -12400/1520 eV
E1 = -8.16 eV is the ground state energy
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E2 - E1 = 12400/2026 eV
E2 = E1 +12400/2026 eV
E2 = -2.04 eV is the energy of the first excited state
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E3 - E1 = 12400/1709 eV
E3 = -0.902 eV is the energy of the second excited state
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E4 - E1 = 12400/1621 eV
E4 = -0.508 eV is the energy of the third excited state
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E5 - E1 = 12400/1583 eV
E5 = -0.325 eV is the energy of the fourth excited state
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let the longest-wavelength (alpha) lines in the Balmer series = ?
the transcaction from n =3 state to n =2
E3 - E2 = hc/?
? = 12400/(-0.902+2.04) A
=10896.30931 A
let the short-wavelength series limit in the Balmer series for this atom = ?'
E? - E2 = hc/?'
?' = 12400/2.04
=6078.4313A
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