What is ?E for the transition of an electron from n = 7 to n = 4 in a Bohr hydro

Question

What is ?E for the transition of an electron from n = 7 to n = 4 in a Bohr hydrogen atom? What is the frequency, v, of the spectral line produced?

Enter your answer with 4 significant digits.
Enter scientific notation

?E = ____________
(Express your answer for ?E in J.
v =____________
(Express your answer for v in s-1

Explanation / Answer

you have to use the rydberg equation... E / hc = Rh x (1 / n1² - 1 / n2²) where E = energy / photon h = plancks constant = 6.626x10^-34 Js c = speed of light = 3.00x10^8 m/s Rh = rydberg constant for hydrogen n1 = 4 n2 = 3 ************* where Rh = m x e^4 / (8 eo^2 x h^3 x c) and m = rest mass electron = 9.109x10^-31 kg e = 1.602x10^-19 C e0 = permitivity of free space = 8.854x10^-12 C² / N m² h = plancks constant = 6.626x10^-34 kg m² / s c = speed of light = 3.00x10^8 m/s Rh = (9.109x10^-31 kg) x (1.602x10^-19 C)^4 ........______________________________… ........8 x (8.854x10^-12 C² / N m²)² x (6.626x10^-34 kg m² / s)^3 x 3.00x10^8 m/s Rh = 1.097x10^7 m^-1 fyi.. watch the units when doing this calculation... in terms of units... kg C^4 / [(C^2 / (kg m /s^2) m)^2 x (kg m² / s)^3 x (m/s)] --> m^-1 ************** and finally... E = (6.626x10^-34 J s) x (3.00x10^8 m/s) x (1.097x10^7 /m) x (1 / 16 - 1 / 9) = -1.06x10^-19 J and that is per atom...the - sign means that is energy released. in terms of per mole... -1.06x10^-19 J / atom x (6.022x10^23 atoms / mole) = -63.4 kJ / mole

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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