Respuesta :
Answer: [tex]0.81\times 10^{16}[/tex] beta particles
Explanation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass = 14.0 g
Molar mass = 137 g/mol
[tex]\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles[/tex]
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
1 mole of cesium contains atoms = [tex]6.023\times 10^{23}[/tex]
0.102 moles of cesium contains atoms = [tex]\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}[/tex]
The relation of atoms with time for radioactivbe decay is:
[tex]N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}[/tex]
Where [tex]N_t[/tex] =atoms left undecayed
[tex]N_0[/tex] = initial atoms
t = time taken for decay = 3 minutes
[tex]{t_{\frac{1}{2}}}[/tex] = half life = 30.0 years = [tex]1.577\times 10^7[/tex] minutes
The fraction that decays : [tex]1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}[/tex]
Amount of particles that decay is = [tex]0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}[/tex]
Thus [tex]0.81\times 10^{16}[/tex] beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.
