The observed lines in the emission spectrum of atomic hydrogen are given by: v(c

Question

The observed lines in the emission spectrum of atomic hydrogen are given by: v(cm^-1) = RH(cm^-1) (1/n^2_1 - 1/n^2_2) (n_2>n_1). The Lyman, Balmer and Paschen series refer to n1 = 1, 2 and 3, respectively. (a) Calculate the highest possible energy of a photon that can be observed in the emission spectrum of Hydrogen. This would correspond to emission from an electron in a shell with a very large principle number (like infinity). [2.18 x 10^-18 J] (b) Calculate the longest and the shortest wavelength observed in the Balmer series. Draw the corresponding energy diagram for each transition. [lambda = 658 nm and 365 nm]

Explanation / Answer

a) n1 = 1 , n2 = infinity which means 1/n2 = 0

hence wavenumber = Rh ( 1/n1^2) = Rh = 10973731.6 m-1

E = ( h x c x wavenumber) = ( 6.625x10^-34 x 3x10^8 x 10973731.6) = 2.18 x 10^ -18 J

b) Balmer n1 = 2 , n2 = 3

Then waveumber = ( 10973731.6) ( 1/2^2 -1/3^2) = 1524129.4 m-1

wavelength = 1/ wavenumber = ( 1/1524129.4) = 6.56 x 10^ -7 m = 656 x 10^-9 m = 656 nm   is longest wavelength

shortes wavelength will be when n2 = infinite , i.e 1/n2 = 0

then wavenumber = 10973731.6 ( 1/2^2-0) = 2743432.9 m-1

Corresponding diagram is just showing lines of transition

One is from energy level n = 3 to n = 2

other is showing from infinity ( i.e outside) to n = 2

wavelength = ( 1/2743432.9) = 3.65 x 10^ -7 m = 365 x 10^ -9 m= 365 nm

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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