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InterviewSolution
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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» (a) For element A Ground state energy, `E_(1) = -2eV` Excited state energy, `E_(2) = 0 eV` Energy of photon emitted, `E = E_(2) - E_(1)` `= 0 - (-2) = 2 eV` `:.` Wavelength of photon emitted, `lamda = (hc)/(E) = (6.626 xx 10^(-34) xx 3xx 10^(8))/(2 xx 1.6 xx 10^(-19)) = (19.878 xx 10^(-7))/(3.2) = 6.211 xx 10^(-7) m = 621 : 1nm` For element B `E_(1) =4.5 eV, E_(2) = 0eV` `E = 0 - (-4.5) = 4.5 eV` `:. lamda = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(4.5 xx 1.6 xx 10^(-19)) = (19.878 xx 10^(-7))/(7.2) = 2.760 xx 10^(-7) = 276 nm` For element C `E_(1) = -4.5 eV, E_(2) = -2 eV` `E = -2- (-4.5) = -2 + 4.5 = 2.5 eV` `:. " " lamda = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(2.5 xx 1.6 xx 10^(-19))` `= (19.878 xx 10^(-7))/(4) = 4.969 xx 10^(-7) m = 496.9 nm` For element D `E_(1) = -10 eV, E_(2) = -2eV` `E = -2-(-10) = 8 eV` `:. lamda (6.626 xx 10^(-34) xx 3 xx 10^(8))/(8xx 1.6 xx 10^(-19)) = (19.878 xx 10^(-7))/(1.28) = 1.552 xx 10^(-7) m = 155.2 nm` `:. Element B has a photon of wavelength 275 nm (b) Element A has radiation of maximum wavelength 621 nm |
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