Given that the mass of an average linebacker at Ursinus College is 250 lbs and the radius of a pea is 0.50 cm, calculate the number of linebackers that would be required to be stuffed into the volume of a pea in order to obtain the same density as an alpha particle.

1.544*10⁹ Linebackers would be required in order to obtain the same density as an alpha particle

Step-by-step explanation:

Assuming that the pea is spherical (with radius R = 0.5 cm = 0.005 m), then its volume is

V = 4/3π*R³ = 4/3π*R³ = 4/3*π * (0.005 m) ³ = 5.236*10⁻⁷ m³

the mass in that volume would be m = N*L (L = mass of linebackers=250Lbs = 113.398 Kg)

The density of an alpha particle is ρa = 3.345*10¹⁷ kg/m³ and the density in the pea ρ will be

ρ = m/V

since both should be equal ρ=ρa, then

ρa = m/V = N*L/V → N = ρa*V/L

replacing values

N = ρa*V/L = 3.345*10¹⁷ kg/m³ * 5.236*10⁻⁷ m³ / 113.398 Kg = 1.544*10⁹ Linebackers

N=1.544*10⁹ Linebackers