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The velocity of `e^(-)` in a certain Bohr orbit of the hydrogen atom bears the ratio `1:275` to the velocity of light. What is the quantum no. "n" of the orbit and the wave no. of the radiation emitted for the transition form the quantum state `(n+1)` to the ground state. |
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Answer» Correct Answer - `2;9.75xx10^(4)cm^(-1)` `V_(n)=2.18xx10^(6)xx(Z)/(n)=(2.18xx10^(6))/(n)` `(2.18xx10^(6))/(n)=(1)/(275)` `(2.18xx10^(6))/(n)=(1)/(3xx10^(8))=(1)/(275)` `(2.18)/(n(300))=(1)/(275)" "(1)/(n)=(300)/(599.5)` `n=(599.5)/(300)=(1)/(275)" "(1)/(n)=(300)/(599.5)` `n=1.99~=2` |
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