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If the value of `E=-78.4 "kcal//mol"`, the order of the orbit in hydrogen atom is-A. `4`B. `3`C. `2`D. `1` |
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Answer» Correct Answer - C `E=-78.4 kcal//mol` `E_(n)=-313.6xx(Z^(2))/(n^(2))kcal//mol` for `H` atoom `Z=1 " "-78.4=313.6xx(1)/(n^(2))` `n^(2)=(313.6)/(78.4)" "n=2` |
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