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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. `1.99`B. `2.14`C. `0.50`D. `0.48` |
Answer» Correct Answer - B `sqrt(C/lambda)=a (Z-b)` `b=1 sqrt(lambda_(Cu)/lambda_(Me))=((Z_(Me)-1)/(Z_(Cu)-1))` `lambda_(Cu)/lambda_(Me)=(41/28)^(2)=2.14` |
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