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The electron in the hydrogen atom makes a transition from the `n = 2` energy level (or state) to the ground state (corresponding to `n = 1`). Find the frequency of the emitted photon. Strategy: Use Eq. directly to obtain `bar(v)`, with `n_(f) = 1` (ground state) and `n_(i) = 2`. Then find the frequncy. |
Answer» Step `1: bar(v) = R_(H) |((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))| = R_(H)|((1)/(1^(2))-(1)/(2^(2)))|` `= (3)/(4)R_(H) = (3)/(4) (1.097 xx 10^(7)m^(-1))` `=0.8228 xx 10^(7)m^(-1) = 8.228 xx 10^(6)m^(-1)` This wavenumber lies in the ultraviolet region. Step `2:` Since `v = cbar(v)`, the frequency of the emitted photon is `v = cbar(v)` `(3.0xx10^(8)ms^(-1))(8.228xx10^(6)m^(-1))` `= 24.68 xx 10^(14) s^(-1) = 2.5xx10^(15)s^(-1) (or Hz)` |
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