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

`v=1.043* 10^(-9)\ m/s`

Step-by-step explanation:

The expression for kinetic energy is:

`K.E.=\frac {1}{2}* m* v^2`

K.E. is the kinetic energy =
`2.176* 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* 10^(-21)=(1)/(2)* 0.004* v^2`

`v^2=(2.176* 10^(-21)* 2)/(0.004)`

`v=\sqrt{(4.352)/(10^(21)* \:0.004)}`

`v=1.043* 10^(-9)\ m/s`