Question

A helium atom has a rest mass of *m*He=4.002603u. When
disassembled into its constituent particles (2 protons, 2 neutrons,
2 electrons), the well-separated individual particles have the
following masses: *m*p=1.007276u, *m*n=1.008665u,
*m*e=0.000549u.

How much work is required to completely disassemble a helium
atom? (*Note:* 1 u of mass has a rest energy of 931.49
MeV.)

Answer #1

This is about the famous Einstein equation, E = m*c^2.

In the case at hand, the sum of the rest masses of the individual particles is

2*(1.007276+1.008665+0.000549) = 4.03298

Where did the "extra" mass come from, when the helium atom was pulled apart?

It is the mass equivalent of the energy added. The work that was done to disassemble the atom is

E = (change in m)*c^2, where c is the speed of light.

E = (0.03038 u)(1.6605*10^(-27) kg/u)(2.9979*10^8 m/s^2

= 4.545 * 10^(--11) kg*m^2 / s^2 = 45.45 picojoules.

For the balloon, let's choose a smallish size like a party balloon, volume maybe 2 or 3 liters.

At STP it might hold a tenth of a mole of helium.

A mole is roughly 6 * 10^23 molecules (or in this case, atoms).

45 picojoules times 6*10^22 is 2.7*10^24 picojoules = 2.7 * 10^12 joules or 2.7 TJ.

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