InterviewSolution
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InterviewSolution
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if `lambda_(Cu)` is the wavelength of `K_alpha`, X-ray line fo copper (atomic number 29) and `lambda_(Mo)` is the wavelength of the `K_alpha` X-ray line of molybdenum (atomic number 42), then the ratio `lambda_(Cu)/lambda_(Mo)`is close to (a) 1.99 (b) 2.14 (c ) 0.50 (d) 0.48 |
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Answer» Correct Answer - B `K_alpha transition takes place from n_1 = 2 to n_2 =1` `:. (1)/(lambda) = R(Z - b)^2 [(1)/((1))^2-(1)/((2))^2)]` For K-series, b=1 `:. (1)/(lambda) prop (Z-1)^2` `rArr (lambda_(Cu))/(lambda_Mo)) = ((Z_(mo) -1)^2/((Z_(Cu) - 1)^2 = (42-1)^2/(29-1)^2` `=(41xx41)/(28xx28) = (1681)/(784) = 2.144` |
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