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The energy levels of an atom are as shown in figure . Which one of those transition will result in the emission of a photon of wavelength `275 nm`? |
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Answer» Here, `lambda = 275 nm = 275 xx 10^(-9) m` Therefore, energy of the emitting photon, `E = (hc)/(lambda) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(275 xx 10^(-9))` `= 7.22 xx 10^(-19) J = (7.22 xx 10^(-19))/(1.6 xx 10^(-19)) = 4.5 e V` The photon of energy `4.5 e V` ("or of wavelength 275 nm")` will be emitted , corresponding to the transition `B`. |
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