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

• A. \(\frac{9}{16}\ E_2\)
• B. \(\frac{16}{9}\ E_2\)
• C. \(\frac{3}{16}\ E_2\)
• D. \(\frac{16}{3}\ E_2\)

Answer 1

The Correct Option is B

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:

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

Answer 2

Step 1: Energy of Electron Formula

$E_n = -13.6 \frac{Z^2}{n^2} eV$

Step 2: Energy of Electron in 2nd Orbit of Hydrogen ($E_2$)

For Hydrogen, $Z=1$, $n=2$

$E_2 = -13.6 \frac{1^2}{2^2} = -13.6 \frac{1}{4} eV$

Step 3: Energy of Electron in 3rd Orbit of He⁺ ($E_3^{He+}$)

For He⁺, $Z=2$, $n=3$

$E_3^{He+} = -13.6 \frac{2^2}{3^2} = -13.6 \frac{4}{9} eV$

Step 4: Express $E_3^{He+}$ in terms of $E_2$

From Step 2, $-13.6 = 4 E_2$

Substitute this into Step 3:

$E_3^{He+} = (4 E_2) \frac{4}{9} = \frac{16}{9} E_2$

Final Answer: The final answer is Option (B) is $\frac{16}{9} E_2$.