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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 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.

Sagot :

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 .