Question

One mole of helium atoms has a mass of $\mathbf{4}$grams. If a helium atom in a balloon has a kinetic energy of$\mathbf{1}\mathbf{.}\mathbf{437}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}}\mathit{J}$, what is the speed of the helium atom? (The speed is much lower than the speed of light.)?

Short answer

The speed of helium atom is,$657.9m/s$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

- The mass of one mole of helium is, $m_m=4\mathrm{g}=4\times {10}^3\mathrm{kg}$

- The kinetic energy of helium atom is, $K=1.437\times {10}^{21}\mathrm{J}$

Significance of kinetic energy

Kinetic energy is the energy the body retains due to motion. Every moving object has kinetic energy.

Determination of the kinetic energy in +x direction

The mass of one mole of helium atom is given, $m_m=4\mathrm{g}=4\times {10}^3\mathrm{kg}$. The $1$mole has $6.022\times {10}^{23}$atom. Then,

The mass of $1$atom is expressed as

$=\frac{4\times {10}^{-3}}{6.022\times {10}^{23}}\mathrm{kg}$

$=6.64\times {10}^{-27}\mathrm{kg}$

The relation of kinetic energy is expressed as,

$K-\frac{1}{2}m{v}^2$

Here $K$is the kinetic energy, $m$is the mass of the helium atom and $V$is the speed of the helium atom.

Substitute $6.64\times {10}^{-27}\mathrm{kg}$for $\mathrm{m}$and $1.437\times {10}^{-21}\mathrm{J}$for $K$in the above equation.

$1.437\times {10}^{-21}\mathrm{J}=\frac{1}{2}\times 6.64\times {10}^{-27}\mathrm{kg}\times {\mathrm{v}}^2$

$v=657.9\mathrm{m}/\mathrm{s}$

Hence the speed of helium atom is, $657.9\mathrm{m}/\mathrm{s}$