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NEET PHYSICSATOMSMedium

Question

The ground state energy of hydrogen atom is 13.6 eV. The energy needed to ionise the atom from its second excited state is:

A

1.51 eV

B

3.4 eV

C

13.6 eV

D

12.1 eV

Step-by-Step Solution

The energy of an electron in the nn-th orbit of a hydrogen atom is given by En=13.6n2E_n = -\frac{13.6}{n^2} eV . The ground state corresponds to n=1n=1. The 'second excited state' corresponds to the principal quantum number n=3n=3 (as n=2n=2 is the first excited state).

Calculating the energy of the electron in the n=3n=3 orbit: E3=13.632 eV=13.69 eV1.51 eVE_3 = -\frac{13.6}{3^2} \text{ eV} = -\frac{13.6}{9} \text{ eV} \approx -1.51 \text{ eV}.

The ionization energy is the energy required to remove the electron from this state to infinity (where energy is 0). Ionization Energy = EE3=0(1.51) eV=+1.51 eVE_{\infty} - E_3 = 0 - (-1.51) \text{ eV} = +1.51 \text{ eV}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ATOMS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSATOMSgroundenergyhydrogenenergyneeded

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