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# A 12.5eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Last updated date: 13th Jun 2024
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Hint: Due to the external energy from the beam, the hydrogen atom in the ground state gets excited to higher levels. The wavelength of the light emitted will depend on the energy differences of the levels of hydrogen atom.
Formula used:
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.

Complete answer: or Complete step by step answer:
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.