Ask Question
4 June, 18:36

An electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ = 93.8 nm. find the principal level to which the electron relaxed.

+2
Answers (1)
  1. 4 June, 18:55
    0
    We can find the lower level by using Rydberg's formula: 1/w = R (1/L² - 1/U²) where:

    w - wavelength (9.38nm),

    L - lower energy level

    U - upper energy level (6); and

    R - Rydberg's constant (10,967,758 waves per meter for hydrogen).

    So plugging in the information above:

    1 / (9.38 * 10**-8 m.) = 10,967,758 (1/L² - 1-36)

    But we need to solve for L.

    The left side (1/9.38 * 10**-8) becomes 10,660,980

    So translating:

    10660980 = 10967758 (1/L² - 0.02777777)

    Then divide both sides by 10967758

    we get: 0.9720291 = 1/L² - 0.02777777.

    then add 0.02777777 to both sides

    we get: 0.9998 = 1/L²

    0.9998 is very close to 1, so we can approximate 1/L² = 1 so L = 1.

    Therefore the lower level is 1 which is the ground state.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “An electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ = 93.8 nm. find the principal ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers