Question

$1.00\times {10}^6$sodium atoms are excited to the $3p$state at $t=0s$. At what time have $8.0\times {10}^5$photons been emitted ?

Short answer

At $27ns$,$8.0\times {10}^5$photons been emitted.

Step-by-step solution

Given Information

We have given that,

$1.00\times {10}^6$sodium atoms are excited to $3p$state at $t=0s.$

We have to find the time at which$8.0\times {10}^5$photons have emitted.

Simplification

From the table given below the sodium $3p$state has a lifetime of $\tau =17ns.$

Some excited-state lifetimes

Atom	State	Lifetime(ns)
Hydrogen	2p	1.6
Sodium	3p	17
Neon	3p	20
Potassium	4p	26

We know that, the number of excited atoms is given by

$N_{exc}=N_0{e}^{-\frac{t}{\tau }}$($N_0$= No. of nuclei at $t=0$,$\tau =$time constant)

Every atom that transitions from the excited state to the ground state emits a photon, so the number $N$of photons emitted is $N=N_0-N_{exc}=8.0\times {10}^5.$. Therefore, the time at which $N$photons have been emitted is

$N=N_0-N_{exc}=N_0(1-e^{-t/\tau })$

$t=\tau \ln \left(\frac{N_0}{N_0-N}\right)=\left(17ns\right)\ln \left(\frac{1.00\times {10}^6}{1.00\times {10}^6-0.8\times {10}^6}\right)=27ns$

At localid="1650351508107" $27ns,8.0\times {10}^5$been emitted.