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An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, exciting the atom
Question
An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, exciting the atom to the highest energy level allowed. List all the possible quantum-jump transitions by which the excited atom could emit a photon and the wavelength (in nm) of each.
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2021-08-12T21:16:34+00:00
2021-08-12T21:16:34+00:00 1 Answers
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Answer:
Explanation:
An electron accelerates through a 12.5 V potential difference, starting from rest, so it will acquire kinetic energy of 12.5 eV .
In hydrogen atom energy of n th orbit in terms of eV is given as follows
En = -13.6 / n² eV
Total energy of 1 st orbit E₁ = – 13.6 eV
Total energy of 2 st orbit E₂ = – 13.6 eV / 2² = – 3.4 eV
Total energy of 3 st orbit E₃ = – 13.6 eV / 3² = – 1.5 eV
Total energy of 4 st orbit E₄ = – 13.6 eV / 4² = – 0.85 eV
E₄ – E₁ = 13.6 – 0.85 = 12.75 eV
E₃ – E₁ = 13.6 – 1.5 = 12.10 eV
E₂ – E₁ = 13.6 – 3.4 = 10.2 eV .
The electron has energy of 12,5 eV so it can excite electron from E₁ to E₃ . .
Jump possible = E₃ to E₂ , E₂ to E₁ and E₃ to E₁
Energy of E₃ to E₂ = 3.4 – 1.5 eV = 1.9 eV
wavelength = 1237 / 1.9 nm = 651 nm
E₃ – E₁ = 13.6 – 1.5 = 12.10 eV
wavelength = 1237 / 12.10 nm = 102.23 nm
E₂ – E₁ = 13.6 – 3.4 = 10.2 eV
wavelength = 1237 / 10.2 nm = 121.27 nm
wavelength of photon possible are 651 nm , 121.27 nm , 102.23 nm .