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.)

Respuesta :

Answer:

[tex]v=1.043\times 10^{-9}\ m/s[/tex]

Explanation:

The expression for kinetic energy is:

[tex]K.E.=\frac {1}{2}\times m\times v^2[/tex]

K.E. is the kinetic energy = [tex]2.176\times 10^{-21}\ J[/tex]

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

v is the velocity

Applying values as:

[tex]2.176\times 10^{-21}=\frac{1}{2}\times 0.004\times v^2[/tex]

[tex]v^2=\frac{2.176\times 10^{-21}\times 2}{0.004}[/tex]

[tex]v=\sqrt{\frac{4.352}{10^{21}\times \:0.004}}[/tex]

[tex]v=1.043\times 10^{-9}\ m/s[/tex]