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Q.
If the energy of a hydrogen atom in nth orbit is $ \Rightarrow $ then energy in the nth orbit of a singly ionized helium atom will be
BVP MedicalBVP Medical 2007
Solution:
Key Idea: Atomic number of helium is twice that of hydrogen. When e, m are the charge and mass of electron in the nth orbit, h is Plancks constant and Z is the atomic number, then the energy of the electron in the electron in the nth orbit is $ E=-\frac{M\,{{Z}^{2}}\,{{e}^{4}}}{8{{\varepsilon }_{o}}{{h}^{2}}}.\frac{1}{{{n}^{2}}} $ Given, $ {{Z}_{H}}=1,\,\,{{Z}_{He}}=2 $ $ \therefore $ $ \frac{{{E}_{H}}}{{{E}_{He}}}=\frac{Z_{H}^{2}}{Z_{He}^{2}}=\frac{1}{4} $ $ \Rightarrow $ $ {{E}_{He}}=4{{E}_{H}} $ Given, $ {{E}_{H}}={{E}_{n}} $ $ {{E}_{He}}=4{{E}_{n}} $