Using the Schrödinger equation for the hydrogen atom, the ground‑state energy is $E_1=-\frac{m_e{e}^4}{8{\epsilon }_0^2{h}^2}$. Given $m_e=9.11\times 10^{-31},\text{kg}$, $e=1.602\times 10^{-19},\text{C}$, ${\epsilon }_0=8.854\times 10^{-12},{\text{C}}^2!!/!{\text{N⋅m}}^2$, and $h=6.626\times 10^{-34},\text{J⋅s}$, calculate $E_1$ in electron‑volts.

Step-by-step solution

1. Compute the numerator: $m_e{e}^4=9.11\times 10^{-31},(1.602\times 10^{-19}{)}^4\approx 5.999\times 10^{-106},{\text{J}}^2!s^2$.
2. Compute the denominator: $8{\epsilon }_0^2{h}^2=8(8.854\times 10^{-12}{)}^2(6.626\times 10^{-34}{)}^2\approx 2.751\times 10^{-88},{\text{J}}^2!s^2$.
3. Divide: $E_1=-\frac{5.999\times 10^{-106}}{2.751\times 10^{-88}}\approx -2.18\times 10^{-18},\text{J}$.
4. Convert to eV using $1,\text{eV}=1.602\times 10^{-19},\text{J}$: $E_1\approx -\frac{2.18\times 10^{-18}}{1.602\times 10^{-19}}\approx -13.6\text{eV}$.
5. Thus the ground‑state energy of hydrogen is $-13.6\text{eV}$.

Concept tested: Schrödinger ground‑state energy

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked