Question

At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio \(\frac{{{{\rm{R}}_{\rm{B}}}}}{{{{\rm{R}}_{\rm{A}}}}}\) of their activities after time t itself decays with time t as e^-3t. If the half-life of A is ln2, the half-life of B is:
1. 4ln 2
2. \(\frac{{{\rm{ln\;}}2}}{2}\)
3. \(\frac{{{\rm{ln\;}}2}}{4}\)
4. 2ln 2

Answer 1

Correct Answer - Option 3 : \(\frac{{{\rm{ln\;}}2}}{4}\)

Concept:

Radioactive decay occurs in unstable atomic nuclei that is, ones that don’t have enough binding energy to hold the nucleus together due to an excess protons or neurons.

The activity of the radioactive material is given by the formula:

R = λN

λ = decay constant (s^-1)

N = Number of nuclei of radioactive material

Calculation:

The activity for substance A is given by the formula:

R_A = λ_A N_A

At t = 0, Number of undecayed nuclei (N₀A) = Number of nuclei of radioactive material (N_A)

⇒ R_A = λ_A N_0A

The activity for substance B is given by the formula:

R_B = λ_B N_B

At t = 0, Number of undecayed nuclei (N_0B) = Number of nuclei of radioactive material (N_B)

⇒ R_B = λ_B N_0B

At t = 0, two radioactive substances A and B have equal activities:

λ_A N₀A = λ_B N_0B

\(\Rightarrow \frac{{{{\rm{N}}_{0{\rm{B}}}}}}{{{{\rm{N}}_{0{\rm{A}}}}}} = \frac{{{{\rm{\lambda }}_{\rm{A}}}}}{{{{\rm{\lambda }}_{\rm{B}}}}}\)        ----(1)

The half-life is given by the formula:

\({{\rm{T}}_{1/2}} = \frac{{{\rm{ln\;}}2}}{{\rm{\lambda }}}\) 

From the question, the half-life of A is ln 2.

\(\Rightarrow {{\rm{T}}_{\left( {1/2} \right){\rm{A}}}} = \frac{{{\rm{ln\;}}2}}{{{{\rm{\lambda }}_{\rm{A}}}}}\) 

\(\Rightarrow {\rm{ln\;}}2 = \frac{{{\rm{ln\;}}2}}{{{{\rm{\lambda }}_{\rm{A}}}}}\) 

∴ λ_A = 1        ----(2)      

At t = t, the activity is given as:

\(\frac{{{{\rm{R}}_{\rm{B}}}}}{{{{\rm{R}}_{\rm{A}}}}} = \frac{{{{\rm{\lambda }}_{\rm{B}}}{{\rm{N}}_{\rm{B}}}}}{{{{\rm{\lambda }}_{\rm{A}}}{{\rm{N}}_{\rm{A}}}}}\) 

We know that, N = N₀ e^-λt

\(\Rightarrow \frac{{{{\rm{R}}_{\rm{B}}}}}{{{{\rm{R}}_{\rm{A}}}}} = \frac{{{{\rm{\lambda }}_{\rm{B}}}{{\rm{N}}_{0{\rm{B}}}}{{\rm{e}}^{ - {{\rm{\lambda }}_{\rm{B}}}{\rm{t}}}}}}{{{{\rm{\lambda }}_{\rm{A}}}{{\rm{N}}_{0{\rm{A}}}}{{\rm{e}}^{ - {{\rm{\lambda }}_{\rm{A}}}{\rm{t}}}}}} = {{\rm{e}}^{ - 3{\rm{t}}}}\) 

\(\Rightarrow \frac{{{{\rm{R}}_{\rm{B}}}}}{{{{\rm{R}}_{\rm{A}}}}} = \frac{{{{\rm{\lambda }}_{\rm{B}}}}}{{{{\rm{\lambda }}_{\rm{A}}}}} \times \frac{{{{\rm{\lambda }}_{\rm{A}}}}}{{{{\rm{\lambda }}_{\rm{B}}}}} \times \frac{{{{\rm{e}}^{ - {{\rm{\lambda }}_{\rm{B}}}{\rm{t}}}}}}{{{{\rm{e}}^{ - {{\rm{\lambda }}_{\rm{A}}}{\rm{t}}}}}} = {{\rm{e}}^{ - 3{\rm{t}}}}\) 

On substituting, equation (1) in above equation,

\(\Rightarrow \frac{{{{\rm{R}}_{\rm{B}}}}}{{{{\rm{R}}_{\rm{A}}}}} = \frac{{{{\rm{e}}^{ - {{\rm{\lambda }}_{\rm{B}}}{\rm{t}}}}}}{{{{\rm{e}}^{ - {{\rm{\lambda }}_{\rm{A}}}{\rm{t}}}}}}\) 

\(\Rightarrow \frac{{{{\rm{R}}_{\rm{B}}}}}{{{{\rm{R}}_{\rm{A}}}}} = {{\rm{e}}^{\left( {{{\rm{\lambda }}_{\rm{A}}} - {{\rm{\lambda }}_{\rm{B}}}} \right){\rm{t}}}} = {{\rm{e}}^{ - 3{\rm{t}}}}\) 

On equating power terms,

⇒ (λ_A - λ_B)t = -3t

⇒ λ_A - λ_B = -3

From equation (2),

⇒ λ_B = 3

∴ λ_B = 4

Now, the half-life of B is:

\(\Rightarrow {{\rm{T}}_{\left( {1/2} \right){\rm{B}}}} = \frac{{{\rm{ln\;}}2}}{{{{\rm{\lambda }}_{\rm{B}}}}}\) 

\(\therefore {{\rm{T}}_{\left( {1/2} \right){\rm{B}}}} = \frac{{{\rm{ln\;}}2}}{4}\)