# Why is the total energy of an electron in an atom negative? [duplicate]

Why is the total energy of an electron in an atom negative?

We know that $$E = -\frac{e^2}{8\pi\epsilon_0a_0}.$$ What does the negative sign in the above equation means?

## marked as duplicate by ACuriousMind♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 16 '17 at 12:55

• You can see parallel to a planet moving around the sun. If the energy of the planet is negative, it's moving along the ellipse, if it's positive or zero, it's hyperbolic or parabolic motion. It's easy with planets, because $E = T + V$ and here the potential energy $V$ is stronger than kinetic energy $T$. Remember $V \sim -r^{-1}$. If the body has positive energy, it means it has energy to beat the potential of the sun and fly away. In atoms we don't assume parabolic, hyperbolic or elliptic motions, but the energy idea is the same. – Jimmy Found Apr 16 '17 at 8:42
It simply indicates that, to take the electron from its "orbit" around the nucleus and move it to infinity where the potential energy between the electron and the nucleus is $0$ requires us to supply at least $+E$. If we supply more than $+E$ the electron will have some kinetic energy left when it reaches infinity.