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Question: Calculate the mass of \(14C\) (half life period = 5720 years) atoms which give \(3.7 \times 10^{7}\)...

Calculate the mass of 14C14C (half life period = 5720 years) atoms which give 3.7×1073.7 \times 10^{7} disintegrations per second

A

2.34×10−4g2.34 \times 10^{- 4}g

B

2.24×10−4g2.24 \times 10^{- 4}g

C

2.64×10−4g2.64 \times 10^{- 4}g

D

2.64×10−2g2.64 \times 10^{- 2}g

Answer

2.24×10−4g2.24 \times 10^{- 4}g

Explanation

Solution

Let the mass of 14C14C atoms be m g

Number of atoms in m g of14C=m14×6.02×102314C = \frac{m}{14} \times 6.02 \times 10^{23}

λ=0.693Half life=0.6935720×365×24×60×60=3.84×10−12sec⁡−1\lambda = \frac{0.693}{\text{Half life}} = \frac{0.693}{5720 \times 365 \times 24 \times 60 \times 60} = 3.84 \times 10^{- 12}\sec^{- 1}{}We know that −dNtdt=λ⋅Nt- \frac{dN_{t}}{dt} = \lambda \cdot N_{t}

i.e. Rate of disintegration =λ×No. of atoms= \lambda \times \text{No. of atoms}

}{= \frac{3.84 \times 10^{- 12} \times m \times 6.02 \times 10^{23}}{14} = 2.24 \times 10^{- 4}g}$$