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The wavelength of the radiation emitted when the electron jumps from 4th shell to 2nd shell isA. `4862 Å`B. `2056Å`C. `5241 Å`D. `109700` cm |
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Answer» Correct Answer - A According to Balmer equation Wave number `(barv)=109677(1/n_(1)^(2)-1/n_(2)^(2))cm^(-1)` `barv=109677(1/((2)^(2))-1/((4)^(2)))cm^(-1)` `=(109677xx3)/16cm^(-1)` `lambda=1/v=16/(109677xx3)=4862xx10^(-8)`cm `=4862xx10^(-10)m=4862 Å` |
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