InterviewSolution
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InterviewSolution
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The ionization energy of a hydrogen like Bohr atom is `4` Rydberg. If the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state is `N-m` and if the radius of the first orbit of this atom is `r-m` then the value of `(N)/(r )=Pxx10^(2)` then, value of `P`. (Bohr radius of hydrogen `=5xx10^(11)m,1` Rydberg `=2.2xx10^(16)J`) |
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Answer» Correct Answer - `12` (a) In energy units, `1` rydberg `=13.6 eV`. The energy needed to detach the electron is `4xx13.6 eV`. The energy in the ground state is, therefore, `E_(1) = -4xx13.6 eV`. The energy of the first excited state `(n=2)` is `E_(2)=(E_(1))/(4)=13.6 eV. DeltaE=40.8 eV`. The wavelength of the radiation emitted is `lambda=(hc)/(DeltaE)` (b) The energy of a hydrogen-like ion in ground state is `E=Z^(2)E_(0)` where `Z=` atomic number and `E_(0) = -13.6 eV`. Thus, `Z=2`. The radius of th first orbit is `(a_(0))/(Z)` where `a_(0)=5xx10^(-11)m`. Thus, `r=(a_(0))/(Z)=2.5xx10^(-11)m` |
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