What is the energy (in joules) of an electron in a helium ion (He+) at the energy level n = 4? (Use the formula E = -Z^2 R_H / n^2, where Z is the atomic number, R_H = 2.18 × 10^{-18} J, and n is the principal quantum number.)

• A. -1.36 × 10^{-19} J
• B. -6.81 × 10^{-20} J
• C. -2.18 × 10^{-18} J
• D. -8.72 × 10^{-19} J

Answer: (B)

Verified step by step guidance

1. Identify the given values from the problem: the atomic number \(Z\) for helium is 2, the Rydberg constant \(R_H\) is \$2.18 \times 10^{-18}\( J, and the principal quantum number \)n$ is 4.
2. Write down the formula for the energy of an electron in a hydrogen-like ion: \(E = -\frac{Z^2 R_H}{n^2}\)
3. Substitute the known values into the formula: replace \(Z\) with 2, \(R_H\) with \$2.18 \times 10^{-18}\( J, and \)n$ with 4.
4. Calculate the numerator by squaring the atomic number: \(Z^2 = 2^2 = 4\), then multiply by \(R_H\) to get \$4 \times 2.18 \times 10^{-18}$ J.
5. Calculate the denominator by squaring the principal quantum number: \(n^2 = 4^2 = 16\), then divide the numerator by this value to find the energy \(E\).