The electron, in a hydrogen atom, initially in a state of quantum number $$n1$$, makes a transition to a state whose excitation energy, with respect to the ground state, is $$10.2eV$$. If the wavelength, associated with the photon emitted in this transition, is $$487$$ $$.5$$ nm, find the value of the quantum number, $$n1$$ of the electron in its initial state.

Question

Infinity Learn · Accepted Answer

Energy of an electron in the nth orbit of $$H-atom$$,$$En=$$−$$13.6n2eVE1=$$−$$13.6eV$$,$$E2=$$−$$3.4eV$$,$$E3=$$−$$1.51eV$$,… Clearly, $$E2$$−$$E1=$$−$$3.4$$−$$($$−$$13.6)=10.2eVThus$$ the energy state $$n=2$$ has an excitation energy of $$10.2$$ eV with respect to the ground state. Hence the electron is making a transition from state n $$=$$ $$n1$$ to the state $$n=2$$, where $$n1$$ > $$2$$. Now, $$En1$$−$$E2=hc$$λ$$=124004875eV=2.55eVEn1=E2+2.55eV=$$−$$3.4+2.55=$$−$$0.85eV$$ Also, $$En1=$$−$$13.6n12eV$$∴−$$13.6n12=$$−$$0.85$$⇒       $$n1=13.60.85=16=4$$

Bohr's atom model

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