Question

One mole of helium atoms has a mass of 4 grams. If a helium atom in a balloon has a kinetic energy of 2.176 × 10^-21J, what is the speed of the helium atom? (The speed is much lower than the speed of light.)

Answer 1

Answer:

$v=1.043\times 10^{-9}\ m/s$

Explanation:

The expression for kinetic energy is:

$K.E.=\frac {1}{2}\times m\times v^2$

K.E. is the kinetic energy = $2.176\times 10^{-21}\ J$

m is the mass  = 4 g = 0.004 kg (As 1 g = 0.001 kg)

v is the velocity

Applying values as:

$2.176\times 10^{-21}=\frac{1}{2}\times 0.004\times v^2$

$v^2=\frac{2.176\times 10^{-21}\times 2}{0.004}$

$v=\sqrt{\frac{4.352}{10^{21}\times \:0.004}}$

$v=1.043\times 10^{-9}\ m/s$