Questions & Answers

Question

Answers

Answer

Verified

130.8k+ views

The formula to obtain the wavelengths for the transitions between levels of hydrogen atom is given by:

$\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n^2}-\dfrac{1}{p^2}\right)$

Where, $R_H$ is Rydberg constant with a value of around $1.09\times10^{-7} m^{-1}$ for hydrogen atom.

The nth energy level of a hydrogen atom has energy given by:

$E=\dfrac{-13.6}{n^2}$ eV

Therefore, the difference between the energies of two levels can be found with the help of this formula.

The difference between the energies of level n=1 and level n=2 is given as:

$E _2 –E_1=-13.6\left(\dfrac{1}{2^2}-\dfrac{1}{1^2}\right)=10.2$ eV

Similarly, difference between the energies of level n=3 and level n=1 is given as:

$E _3 –E_1=-13.6\left(\dfrac{1}{3^2}-\dfrac{1}{1^2}\right) \approx 12.09$eV

Thus, in the question, we are giving 12.5eV energy to the hydrogen atom. This amount of energy is sufficient to excite the hydrogen atom to level with n=3.

Thus, we will get 3 wavelengths corresponding to following transitions:

(i)n=3 to n=1.

(ii)n=3 to n=2.

(iii)n=2 to n=1.

: The formula to obtain the wavelengths for the transitions between levels of hydrogen atom is given by:

$\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n^2}-\dfrac{1}{p^2}\right)$

Where, $R_H$ is Rydberg constant with a value of around $1.09\times10^{-7} m^{-1}$ for hydrogen atom.

Using this formula, we may write:

(i)n=1 and p=3, gives us $\lambda$=103nm approximately.

(ii)n=2 and p=3, gives us $\lambda$=696nm approximately.

(iii)n=1 and p=2, gives us $\lambda$=122nm approximately.

Transitions to level 2 from any level fall under Balmer series and transitions to level 1 from any level fall under Lyman series

The values of n and p upon switching might produce negative wavelengths here. There are 3 transitions in total but one might conclude 2 transitions ignoring the 3 to 2 transition.