The shortest wavelengths emitted in hydrogen spectrum corresponding to different spectral series are as under:

(A) Pfund series
(B) Balmer series
(C) Brackett series
(D) Lyman series

The wavelengths arranged correctly in decreasing order are________.

Fill in the blank with the correct answer from the options given below

The correct option is (A), (C), (B), (D).

To find the correct decreasing order of the shortest wavelengths emitted in the hydrogen spectrum for different spectral series, we can use the Rydberg formula for the wavelength of emitted radiation:
$$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
where $R$ is the Rydberg constant, $n_1$ is the lower energy level, and $n_2$ is the higher energy level of the electronic transition.

The shortest wavelength in any spectral series corresponds to the transition of maximum energy (also known as the series limit), which occurs when an electron falls from the outermost boundary level, i.e., $n_2=\infty$.
Substituting $n_2=\infty$ into the Rydberg formula gives:
$$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-0\right)=\frac{R}{n_1^2}$$
Simplifying for the wavelength $\lambda$, we obtain:
$$\lambda =\frac{n_1^2}{R}$$

Now, let us calculate the shortest wavelength for each of the given spectral series using their respective ground-state principal quantum number ($n_1$):

1. Lyman series (D):
For this series, $n_1=1$.
The shortest wavelength is:
$${\lambda }_{\text{D}}=\frac{1^2}{R}=\frac{1}{R}$$

2. Balmer series (B):
For this series, $n_1=2$.
The shortest wavelength is:
$${\lambda }_{\text{B}}=\frac{2^2}{R}=\frac{4}{R}$$

3. Brackett series (C):
For this series, $n_1=4$.
The shortest wavelength is:
$${\lambda }_{\text{C}}=\frac{4^2}{R}=\frac{16}{R}$$

4. Pfund series (A):
For this series, $n_1=5$.
The shortest wavelength is:
$${\lambda }_{\text{A}}=\frac{5^2}{R}=\frac{25}{R}$$

Comparing the values obtained above:
$$\frac{25}{R}>\frac{16}{R}>\frac{4}{R}>\frac{1}{R}$$
Therefore, the decreasing order of the wavelengths is:
${\lambda }_{\text{A}}>{\lambda }_{\text{C}}>{\lambda }_{\text{B}}>{\lambda }_{\text{D}}$
This corresponds to the sequence: (A), (C), (B), (D).