Question

$1.0\times {10}^6$sodium atoms are excited to the $3p$state at $t=0$s. How many of these atoms remain in the $3p$state at

(a) $t=10ns$

(b)$t=30ns$, and

(c) $t=100ns$?

Short answer

(a). The atoms is $5.6\times {10}^5$.

(b). The atoms is $1.7\times {10}^5$.

(c). The atoms is$3.0\times {10}^3$.

Step-by-step solution

Part (a) step 1: Given Information

We have to given $t=10ns$.

Part (a) step 2: Simplify

Consider :

$$\begin{gathered} T=17ns \\ {N}_0=1.0\times {10}^6 \end{gathered}$$

using formula:

$N_e=N_0{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{${\rm T}$}\right.}$

Now, we find for $t=10ns$:

$$\begin{gathered} N_e=N_0{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{${\rm T}$}\right.} \\ {N}_e=\left(1.0\times {10}^6\right)e^{\raisebox{1ex}{$-10ns$}\!\left/ \!\raisebox{-1ex}{$17ns$}\right.} \\ {N}_e=5.6\times {10}^5 \end{gathered}$$

Part (b) step 1: Given Information

We have to given $t=30ns$..

Part (b) step 1: Simplify

We find for$t=30ns$.

role="math" localid="1650277679093" $$\begin{gathered} N_e=N_0{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{${\rm T}$}\right.} \\ {N}_e=\left(1.0\times {10}^6\right)e^{\raisebox{1ex}{$-30ns$}\!\left/ \!\raisebox{-1ex}{$17ns$}\right.} \\ {N}_e=1.7\times {10}^5 \end{gathered}$$

Part (c) step 1: Given Information

We have to given $t=100ns$.

Part (c) step 2: Simplify

We find for $t=100ns$.

$$\begin{gathered} N_e=N_0{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{${\rm T}$}\right.} \\ {N}_e=\left(1.0\times {10}^6\right)e^{\raisebox{1ex}{$-100ns$}\!\left/ \!\raisebox{-1ex}{$17ns$}\right.} \\ {N}_e=3.0\times {10}^3 \end{gathered}$$