(c28p47) The Lyman series for a (new!?) one-electron atom is observed in a dista

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 _________________________________________________________________________ _________________________________________________________________________ E - E1 = hc/   where =1520 A0   and E =0      E1 = -12400/1520 eV      E1 = -8.16 eV is the ground state energy _________________________________________________________________________ E2 - E1 = 12400/2026 eV E2 = E1 +12400/2026 eV       E2 = -2.04 eV is the energy of the first excited state ________________________________________________________________________ E3 - E1 = 12400/1709 eV
E3 = -0.902 eV is the energy of the second excited state ________________________________________________________________________ E4 - E1 = 12400/1621 eV        E4 = -0.508 eV is the energy of the third excited state ________________________________________________________________________ E5 - E1 = 12400/1583 eV       E5 = -0.325 eV is the  energy of the fourth excited state _________________________________________________________________________ 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

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.