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InterviewSolution
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In a hypotherical Bohr hydrogen, the mass of the electron is doubled. The energy `E_(0)` and the radius `r_(0)` of the first orbit will be (`a_(0)` is the Bohr radius)A. `E_(0)=-27.eV, r_(0)=a_(0)//2`B. `E_(0)=-27.2 eV, r_(0)=a_(0)=a_(0)`C. `E_(0)=-13.6eV, r_(0)=a_(0)//2`D. `E_(0)=-13.6 eV r_(0)=a_(0)` |
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Answer» Correct Answer - A `E prop m " " and " " rprop(1)/(m)` i.e., energy will becomes two times and radius will become half. |
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