Question

1. We can observe the wavelengths emitted from Hydrogen. When Hydrogen electrons transition between states, they absorb or emit a particle of light called a photon with energy E=hf. Here f is the frequency of light and h is a constant.

a. How much energy does an electron in the n=1 (lowest-energy) state of Hydrogen have? Repeat for n=2 and n=3.

b. How much energy is emitted if an electron in the n=3 state transitions to the n=2 state?

c. Common diffraction gratings have 500 lines/mm and 1000 lines/mm. For the 1000 1 lines/mm grating, what are the angles of emitted light (relative to the central maximum) when an n=3 to n=2 electron transition occurs in Hydrogen?

d. Why don’t we see visible light emitted when hydrogen transitions between the n=2 state and the ground state? Please show all work including how you planned for the problem, did the problem, and a potential way to check your answer!

Answer 1

(a) E,= -13.6 eve - 13.6 eV 12 E2- -13,6 = -3.4ev Ez= -136 --losiev 16) JE = Ez -62 = -1-51+3.4 AE=1.8q ev A E= he a=hl & E 656:08 nm 1240 evnm 1- 89 ev N=1ooo linalom d-I10-3 roma 10-6 m N Noco d Since=na Q = sini nx 656.08x10-9 10-6 Eint(0.656 m) for nel, 0-410 Angles of emitted light na 2 → (4) ** 41- Ez - E, = -3.4 +13.6 = 10. 2 ev d=he = 1240 ev nm 10.2ev =121057 nm GE Sinu, visible range lies lies beth hoonm to foon. We will not see visible light for 121.57hm.