Question

The energy level diagram below roughly shows the lowest four electronic energy levels for a hydrogen atom (ground state E1 =-13.6 eV and the first three excited states: E2, E3 & E4-note that the diagram is not drawn to scale) unbound E4 Es E2 E1

1) If the electron starts out in the ground state and is excited to level E3 by an incoming photon, what was the wavelength of that photon (in nm)?

a) 95.4 nm

b) 102.5nm

c) 121.5nm

d) 136.7 nm

e) 182.3 nm

2) When the electron transitioned from E1 to E3 its orbital radius increased by a factor of:

A) 1 (It didn’t change)

B) 2

C) 3

D) 4

E) 9

3) What is the longest wavelength the hydrogen atom can absorb, if initially in the ground state? Give your answer in nm and explain your reasoning.

Answer 1

1)

electron is excited from level 1 (ground state) to level 3. hence energy absorbed is = E₃ - E₁ = 1.36 ( 1 - 1/9) =

13.6 x 8 /9 eV

$\Delta{E} = \frac{13.6\times8}{9}\times1.6\times10^{-19}=\frac{6.63\times10^{-34}\times3\times10^{8}}{\lambda}$

$\lambda=102.5\,nm$

answer (b)

2)

In Bohr model radius of orbit r is proprotional to square of quantum number n.

$r\propto\,n^2$

when electron makes a transition from E₁ to E₃ radius increases by a factor of 9.

3)

since $\Delta{E}=\frac{hc}{\lambda}$

longest wavelength means smallest energy difference.

That is transition is from ground state to level 2

Energy difference is $\Delta{E}=13.6\left(1-\frac{1}{4} \right )\,eV$

$13.6\times\frac{3}{4}\times1.6\times10^{-19}=\frac{6.63\times10^{-34}\times3\times10^{8}}{\lambda}$

$\lambda=121.5\,nm$