Question

The wavenumber of the spectral line in the emission spectrum of hydrogen will be equal to 8/9 times of the Rydberg’s constant. Then the electron jumps from

• A. 5 → 2
• B. 5 → 3
• C. 3 → 1
• D. 4 → 2

Answer 1

The correct option is C

3 → 1

Explanation for the correct answer:-

Option (C)

Step 1 Calculating the electron's state,

Wave number is defined as the number of waves in a unit distance. It is reciprocal of wavelength (which is length of single cycle in a wave).
The five series are: Balmer, Lyman, Paschen, Bracket and Pfund.
When the spectrum (i.e. the arrangement of different wavelengths in either increasing order or decreasing order) of hydrogen was studied then these five series were found in the spectra.
Rydberg gave the relation between energy difference between different energy levels and wavelength of absorbed or emitted photons is known as Rydberg formula,${\displaystyle \frac{1}{\lambda }}=R{Z}^2[{\displaystyle \frac{1}{{n_i}^2}}-{\displaystyle \frac{1}{{n_h}^2}}]$ where
$\lambda$ is wavelength of the wave
$R$ is Rydberg constant.= $1.097m^{-1}$
$Z$ is atomic number of the element
$n_i$ is lower energy level and $n_h$ is higher energy level.
Lyman series:

By the Rydberg formula if $n_i$ (i.e. the lower energy level) is $1$ and $n_h$ are $2,3,4,..$ , then the series of wavelength will be in Lyman series. If $n_h$ is two then it will represent the first line of Lyman series and if $n_h$ is three then it will represent the second line of Lyman series and so on.

Balmer series:

By the Rydberg formula if $n_i$ (i.e. the lower energy level) is $2$ and $n_h$ are $3,4,5,..$ then the series of wavelength will be in Balmer series.If $n_h$ is three then it will represent the first line of the Balmer series and if $n_h$ is four then it will represent the second line of the Balmer series and so on.

Paschen series:

By the Rydberg formula if $n_i$ (i.e. the lower energy level) is $3$ and $n_h$ (i.e. the higher energy levels) are $4,5,6,..$ , then the series of wavelength will be in Paschen series. If $n_h$ is four then it will represent the first line of the Paschen series and if $n_h$ is five then it will represent the second line of the Paschen series and so on.

Brackett series:

By the Rydberg formula if $n_i$ (i.e. the lower energy level) is $4$ and $n_h$ (i.e. the higher energy levels) are $5,6,7,..$ , then the series of wavelength will be in Brackett series. If $n_h$ is five then it will represent the first line of the Brackett series and if $n_h$ is six then it will represent the second line of the Brackett series and so on.

Pfund series:

By the Rydberg formula if $n_i$ (i.e. the lower energy level) is $5$ and $n_h$ (i.e. the higher energy levels) are $6,7,8,..$ , then the series of wavelength will be in Pfund series. If $n_h$ is six then it will represent the first line of Pfund series and if $n_h$ is seven then it will represent the second line of the Pfund series and so on.

Here we are given that the wave number of the spectral line in the emission spectrum of hydrogen will be equal to ${\displaystyle \frac{8}{9}}$ times the Rydberg’s constant. $n_1=1$

We consider $n_1=1$ because we can see emission spectrum has lower energy $1$ from the given options so we did not need to consider other values of $n$,$Z$=$1$ because Hydrogen has atomic number $1$

${\displaystyle \frac{8}{9}}{R}_H=R_H({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{{n_2}^2}})$ .
On solving, we will get $n_2=3$ . Hence electrons will jump from $n=3$ to $n=1$ .

Hence option C is the correct answer.