Question

Figure shows the variation of the number of radioactive atoms left undecayed with time. The time corresponding to N = $\frac{N_{0}}{3}$is-

• A. 3 t₀
• B. t₀$\log_{e}2$
• C. $\frac{t_{0}\log_{e}3}{\log_{e}\left( \frac{3}{2} \right)}$
• D. Cannot be ascertained because it depends on the decay constant

Answer 1

Answer: (C)

$\frac{t_{0}\log_{e}3}{\log_{e}\left( \frac{3}{2} \right)}$

Answer 2

N = N₀ e^–lt

$\frac { 2 \mathrm {~N} _ { 0 } } { 3 }$ = N₀

$\mathrm { e } ^ { - \lambda \mathrm { t } _ { 0 } }$ =$\frac { 3 } { 2 }$ ........ (1)

lt₀ = log_e $\left( \frac { 3 } { 2 } \right)$

Also $\frac { \mathrm { N } _ { 0 } } { 3 }$ = N₀ $\mathrm { e } ^ { - \lambda t _ { 1 } }$

lt₁ = log_e3

t₁ = $\frac { 1 } { \lambda }$log_e3 =