\(\frac{N_{0}}{3}=N_{0} e^{-\lambda t_{2}}\)      .... \((i)\)

Number of undecayed atom after time \(t_{1};\)

\(\frac{2 N_{0}}{3}=N_{0} e^{-\lambda t_{1}}\)   .... \((ii)\)

From \((i)\), \(e^{-\lambda t_{2}}=\frac{1}{3}\)

\(\Rightarrow \quad-\lambda t_{2}=\log _{e}\left(\frac{1}{3}\right)\)   .... \((iii)\)

From \((ii)\) \(-e^{-\lambda t_{2}}=\frac{2}{3}\)

\(\Rightarrow \quad-\lambda t_{1}=\log _{\mathrm{e}}\left(\frac{2}{3}\right)\)   .... \((iv)\)

Solving \((iii)\) and \((iv)\), we get \(t_{2}-t_{1}=20 \mathrm{\,min}\)