Question

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number nf. Vi and Vf are respectively the initial and final potential energies of the electron. If  VfVi=6.25, then the smallest possible nf is

Solution

Potential energy of a electron in Bohr's modal is given by (assuming Coulombic force) U=−Kze2γ where, γ= radious  & radius for Bohr's orbital (for mono-electronic system) =0.529n2z so, UiUr=γ2γ1=n22n21 ...(1) As, given n2n1=6.25 by comparing above equation to equation (1).  n2n1=√6.25n2n1=2.5 So, n2n1=52 it can be written as,  n2=5, n1=2 so the final value of n2=5. Co-Curriculars

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