p40_047.pdf

47. (a) Since E 2 =

0 . 85eV and E 1 =

−

13 . 6eV+10 . 2eV =

−

3 . 4eV, the photon energy is E photon =

E 2 −

E 1 =

−

0 . 85eV

−

(

−

3 . 4eV)=2 . 6eV .

(b) From

13 . 6eV) 1

=2 . 6eV

E 2 −

E 1 =(

−

n 2 −

n 1

we obtain

n 2 −

n 1

2 . 6eV

13 . 6eV ≈−

16 = 1

2 2

−

4 2 −

Thus, n 2 =4and n 1 = 2. So the transition is from the n =4statetothe n = 2 state. One can

easily verify this by inspecting the energy level diagram of Fig. 40-16.

−