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An element with Z = 11 emits Kalpha-X-ray with wavelength lamda. The atomic no. of element which emits Kalpha X-ray of wavelength 4lamda is: |
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Answer» 6 `v=a^(2)(z-b)^(2)` `:. (C)/(LAMBDA)=a^(2)(z-b)^(2)` `:.(lambda_(1))/(lambda_(2))=((z_(2)-1)^(2))/((z_(1)-z)^(2))` here b=1 put `z_(1)=11, lambda_(1)= lambda, lambda_(2)= 4lambda_(2)` & find `z_(2)` Then `z_(2)=6.` |
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