Question

Place the following transitions of the hydrogen atom in order from longest to shortest wavelength of the photon emitted. Rank from longest to shortest wavelength. To rank items as equivalent, overlap them. n=7 to n=4 n=5 to n=3 n=4 to n=2 n=3 to n=2

Answer 1

Answer:

7 to 4     (higher  $\lambda$)

5 to 3

3 to 2

4 to 2   (lower  $\lambda$)

Explanation:

We can use the Rydberg formula which relates the wavelength of the photon emissions to the principle quantum numbers involved in the transition:

$\frac{1}{\lambda}=R((\frac{1}{n_1^2})-(\frac{1}{n_2^2}))$

with $n_1$ final n, and $n_2$ initial n

evaluating for each transition:

7 to 4 $\frac{1}{\lambda} = R((\frac{1}{4^2})-(\frac{1}{7^2}))= R(1/16-1/49)=R(0.042)$

5 to 3 $\frac{1}{\lambda}= R(1/9-1/25)=R(0.071)$

4 to 2 $\frac{1}{\lambda}= R(1/4-1/16)=R(0.1875)$

3 to 2 $\frac{1}{\lambda}= R(1/4-1/9)=R(0.139)$

Note that the above formula is written for $\frac{1}{\lambda}$, so lower $\frac{1}{\lambda}$ value obtained involves higher $\lambda$.

So we should order from lower to higher $\frac{1}{\lambda}$

7 to 4     (higher  $\lambda$)

5 to 3

3 to 2

4 to 2   (lower  $\lambda$)

Note: Take into account that longer wavelength involves lower energy ($E=\frac{hc}{\lambda}$).