Question

please show work on paper

1.1 (opts). When the electron in a hydrogen atom falls from an excited energy level to the third energy level, a photon with a wavelength of 1005 3 nm is emitted. What is the principle quantum number of the original state?

Answer 1

wavelength = 1005.3 nm

= 1.005*10^-6 m

Here photon will be emitted

1/lambda = R* (1/nf^2 - 1/ni^2)

R is Rydberg constant. R = 1.097*10^7 m-1

1/lambda = R* (1/nf^2 - 1/ni^2)

1/1.005*10^-6 = 1.097*10^7* (1/3^2 - 1/ni^2)

(1/3^2 - 1/ni^2) = 9.068*10^-2

1/ni^2 = 2.043*10^-2

ni^2 = 49

ni = 7

Answer: 7