Question

The energy associated with the first orbit of He+ is:

• A. 0J
• B. -8.72 × 10^-18 J
• C. -4.58 × 10^-18 J
• D. -0.545 × 10^-18 J

Hint:

Use the formula En​=−2.18×10−18×Z2/n2​ for hydrogen-like atoms.

Answer 1

The Correct Option is B

To find the energy associated with the first orbit of \(\text{He}^+\), we can use the formula for the energy levels of the hydrogen-like ion, given by:

\(E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2}\) 

where:

• \(Z\) is the atomic number of the ion.
• \(n\) is the principal quantum number, which denotes the orbit (here, \(n = 1\)).
• The constant value \(13.6\, \text{eV}\) is the ionization energy for hydrogen.

For \(\text{He}^+\), \(\text{Z} = 2\), since helium has an atomic number of 2. Substituting these values into the formula gives:

\(E_1 = -\frac{2^2 \cdot 13.6}{1^2} = -4 \cdot 13.6 = -54.4 \text{ eV}\)

To convert this energy from electron volts to joules, we use the conversion factor \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\).

Therefore,

\(E_1 = -54.4 \cdot 1.602 \times 10^{-19} = -8.713088 \times 10^{-18} \text{ J}\)

Rounding this value, we find:

\(E_1 \approx -8.72 \times 10^{-18} \text{ J}\)

Thus, the energy associated with the first orbit of \(\text{He}^+\) is -8.72 × 10^-18 J, making this the correct answer.

Answer 2

The energy of an electron in the \( n \)-th orbit of a hydrogen-like atom is given by the formula:

\( E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \)

For He⁺, \( Z = 2 \) (atomic number of helium) and \( n = 1 \) (first orbit):

\( E = -2.18 \times 10^{-18} \times \frac{2^2}{1^2} \)

\( E = -2.18 \times 10^{-18} \times 4 = -8.72 \times 10^{-18} \, \text{J} \)

Thus, the energy associated with the first orbit of He⁺ is \( -8.72 \times 10^{-18} \, \text{J} \).