Consider a hydrogen atom. Calculate the frequency f of the n =2 to n =1 transiti

Question

Consider a hydrogen atom. Calculate the frequency f of the n =2 to n =1 transition and compare with the frequency f orb of the electron orbital motionin the n =2 state. Make the same calculations for the n = 10 000 to n = 9999 transition. I get that as the n increases the effect of relativity decreases. I started with E = -13.6/n^2 for n =1 and n=2. Came up with -13.6 eV and -3.4 eV respectfully. However, from there it didn't work otu. I still didn't have the right amount of Hz ( 2.46x 10^15 Hz with a f of 8.31 x 10^14). The larger numbers come out at 6.57 x 10^3 Hz and f = 6.65 Hz. Thank you!

Explanation / Answer

E = E2 - E1 = 13.6*(1/1 - 1/4) = 10.2 eV = 10.2*1.6e-19 J = 1.632e-18 J

f = E/h = 1.632e-18/6.626e-34 = 2.46 x 10^15 Hz

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