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`A` and `B` are two hydrogen like atoms such that `m_(B)=2m_(A)`. Also, the number of protons and neutrons in the two nuclei are equal. Given that difference of photon energy corresponding to the first Balmer lines emitted by `A` and `B` is `2.667eV`. Let `Z_(A)` and `Z_(B)` be the atomic numbers of `A` and `B` respectively.A. `Z_(A)=4`B. `Z_(A)=1`C. `Z_(B)=2`D. `Z_(B)=8` |
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Answer» Correct Answer - B::C `m_(B)=2m_(A)impliesZ_(B)=2Z_(A)`. Let `Z_(A)=Z, Z_(B)=2Z` So, `E_(B)-E_(A)=(13.6xx5)/36xx3Z^(2)=5.667` `implies Z=1` |
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