If a nucleus has a volume of 1.0 × 10^-42 m³, what is its mass number?

Question

Quizplus · Accepted Answer

The volume of the nucleus is given in scientific notation as:

1.0 × 10^-42 m^3

We know that the volume of a sphere is given by the formula:

V = (4/3)πr^3

where V is volume and r is radius.

We can rearrange this formula to solve for the radius:

r = [(3V)/(4π)]^(1/3)

Plugging in the given volume, we get:

r = [(3 x 1.0 × 10^-42)/(4π)]^(1/3)
r = 1.208 x 10^-14 m

Next, we need to use the formula for the volume of a sphere to calculate the volume of the nucleus:

V = (4/3)πr^3

Plugging in the radius we just calculated, we get:

V = (4/3)π(1.208 x 10^-14)^3
V = 9.98 x 10^-42 m^3

Now we can use the fact that the density of a nucleus is approximately constant to calculate the mass of the nucleus:

density = mass/volume

Solving for mass, we get:

mass = density x volume

The density of a nucleus is approximately 2.3 x 10^17 kg/m^3. Plugging this in along with the volume we just calculated, we get:

mass = (2.3 x 10^17 kg/m^3)(9.98 x 10^-42 m^3)
mass = 2.29 x 10^-24 kg

Now we can find the mass number, which is the total number of protons and neutrons in the nucleus. We know that the mass of a proton is approximately 1.0073 atomic mass units (amu), and the mass of a neutron is approximately 1.0087 amu. Therefore:

mass number = (mass of nucleus)/(mass of one proton)

mass number = (2.29 x 10^-24 kg)/(1.0073 amu x 1.66 x 10^-27 kg/amu)
mass number ≈ 138

Therefore, the answer is D) 138.

If a Nucleus Has a Volume of 1