Chapter 41: Q. 22 (page 1207)

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

Short Answer

Expert verified

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

Step by step solution

Given Information

We have given that,

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

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

Simplification

From the table given below the sodium $3 p$ state has a lifetime of $τ = 17 n s .$

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

$N_{e x c} = N_{0} e^{- \frac{t}{τ}}$ ( $N_{0}$ = No. of nuclei at $t = 0$ , $τ =$ 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 $N = N_{0} - N_{e x c} = 8.0 \times 10^{5} .$ . Therefore, the time at which $N$ photons have been emitted is

$N = N_{0} - N_{e x c} = N_{0} (1 - e^{- t / τ})$

$t = τ \ln (\frac{N_{0}}{N_{0} - N}) = (17 n s) \ln (\frac{1.00 \times 10^{6}}{1.00 \times 10^{6} - 0.8 \times 10^{6}}) = 27 n s$

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

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

Short Answer

Step by step solution

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Simplification

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1.00×106sodium atoms are excited to the 3pstate at t=0s. At what time have 8.0×105photons been emitted ?

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Step by step solution

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Simplification

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