Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes

• Matter & Atoms

Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be:

1. 4 : 1
2. 1 : 4
3. 5 : 4
4. 1 : 16

Solution

According to law of radioactive decay:

\( N = N_0 e^{- \lambda t} \) 

Half life is given by:

\( T_{1/2} = \dfrac{\ln 2}{\lambda} \)

\( \lambda_A = \dfrac{\ln 2}{20} \) and \( \lambda_B = \dfrac{\ln 2}{40} \)

After 80 minutes,

\( N_A = N_0 e^{(\ln 2 / 20 ) \times 80} = N_0 e^{-\ln (2^4)} = \dfrac{N_0}{16} \)

\( N_B = N_0 e^{(\ln 2 / 40 ) \times 80} = N_0 e^{-\ln (2^2)} = \dfrac{N_0}{4} \)

Decayed A = \( N_0 - \dfrac{N_0}{16} = \dfrac{15}{16} N_0 \)

Decayed B = \( N_0 - \dfrac{N_0}{4} = \dfrac{3}{4} N_0 \)

Required Ratio = 5/4

The correct option is C.