Chapter 28, Problem 29P

(a)

To determine

The name of all the states $\left(n,l\right)$ to which the atom is allowed to jump with the emission of the photon.

Answer to Problem 29P

The state are $2p,3p,4p,5p,f,5f$ .

Explanation of Solution

Given info:

The excited H atom is in a 6d state.

Formula Used:

Consider as the photon is emitted or absorbed the transition occur only between the states with the values of I that differ by exactly one unit.

  $\Delta I=\pm 1$

Here the atom is allowed to have the jump with the emission of the photon that is from $n=1,2,3,4,5$ for $l=2$ and the new value of l be 1 or 3.

Calculation:

The orbital quantum number is related to the magnitude of the angular moment of the electron. I can take on the integer values from $0$ to $\left(n-1\right)$ , $l=1$ . Correspond s to p subshell and $l=3$ corresponds to $f$ sub shell and the state are $2p,3p,4p,5p,f,5f$

Conclusion:

Therefore, the state are $2p,3p,4p,5p,f,5f$ .

(b)

To determine

The number of different wave length that are there.

Answer to Problem 29P

The number of wavelength is four and the four different photon wavelengths which corresponds to four possible changes in energy.

Explanation of Solution

Calculation:

For energy purposes there are four possible destination states, corresponding to n = 2, 3, 4 and 5.Thus there are four different photon wavelengths which corresponds to four possible changes in energy.

Conclusion:

Therefore, the number of wavelength is four and the four different photon wavelengths which corresponds to four possible changes in energy.