Question:

Energy of an electron in the second orbit of hydrogen atom is E2. The energy of electron in the third orbit of He+ will be

Updated On: Apr 1, 2025
  • \(\frac{9}{16}\ E_2\)
  • \(\frac{16}{9}\ E_2\)
  • \(\frac{3}{16}\ E_2\)
  • \(\frac{16}{3}\ E_2\)
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The Correct Option is B

Solution and Explanation

Given:

  • Energy of electron in 2nd orbit of hydrogen atom: \( E_2 \)
  • Find energy of electron in 3rd orbit of He\(^+\) ion

Step 1: Recall energy level formula for hydrogen-like atoms

The energy of an electron in nth orbit is given by:

\[ E_n = -13.6 \frac{Z^2}{n^2} \text{ eV} \]

where:

  • \( Z \) = atomic number
  • \( n \) = principal quantum number

Step 2: Express given energy \( E_2 \) for hydrogen (Z=1)

For hydrogen's 2nd orbit:

\[ E_2 = -13.6 \frac{1^2}{2^2} = -3.4 \text{ eV} \]

Step 3: Calculate energy for He\(^+\) (Z=2) in 3rd orbit

For He\(^+\)'s 3rd orbit:

\[ E_3^{He^+} = -13.6 \frac{2^2}{3^2} = -13.6 \times \frac{4}{9} \text{ eV} \]

Step 4: Find ratio of energies

We need to express \( E_3^{He^+} \) in terms of \( E_2 \):

\[ \frac{E_3^{He^+}}{E_2} = \frac{-13.6 \times \frac{4}{9}}{-3.4} = \frac{4}{9} \times \frac{13.6}{3.4} = \frac{4}{9} \times 4 = \frac{16}{9} \]

Thus:

\[ E_3^{He^+} = \frac{16}{9} E_2 \]

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