Academic session 2026-27, Chapter 12 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Physics - Chapter Atoms. This page offers a step-by-step solution to the specific question from Exercise 1, Question 9:
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV.
Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.
Orbital energy is related to orbit level (n) as:
E = -13.6/(n)2 eV
For n = 3, E = -13.6/(9)2 = -1.5 eV
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.
During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for Lyman series as:
1/λ = Ry (1/12 - 1/n2)
Where,
Ry = Rydberg constant = 1.097 × 107 m−1
λ= Wavelength of radiation emitted by the transition of the electron
For n = 3, we can obtain λ as:
1/λ = 1.097 x 107 (1/12 - 1/32)
= 1.097x107(1-1/9) = 1.097x107x8/9
λ = 9/(8 x 1.097 x 107) = 102.55 nm
If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:
1/λ = 1.097 x 107 (1/12 - 1/22)
= 1.097x107(1-1/4) = 1.097x107x3/4
λ = 4/(1.097x107x3) = 121.54nm
If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
1/λ = 1.097 x 107 (1/22 - 1/32)
= 1.097x107(1/4-1/9) = 1.097x107x5/36
λ = 36/(5x1.097x107) = 656.33 nm
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 9: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wave....
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