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If `lambda_(Cu)` is the wavelength of `K_(alpha)` X-ray line of 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 toA. 1.99B. 2.14C. 0.5D. 0.48 |
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Answer» Correct Answer - B `(lambda_(Cu))/(lambda_(Mo))=((Z_(Mo)-1)/(Z_(Cu)-1))^(2)=(41xx41)/(28xx28)=(1681)/(784)=2.144` |
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