Cesium-137 undergoes beta decay and has a half-life of 30.0 years. how many beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes?

beta particles

Explanation:

Given mass = 14.0 g

Molar mass = 137 g/mol

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number of particles.

1 mole of cesium contains atoms =  

0.102 moles of cesium contains atoms =  

The relation of atoms with time for radioactivbe decay is:

Where =atoms left undecayed

= initial atoms

t = time taken for decay = 3 minutes

= half life = 30.0 years = minutes

The fraction that decays  :  

Amount of particles that decay is  =

Thus beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.