Consider the atomic system as shown in the figure, where the Einstein A coefficients for spontaneous emission for the levels are \( A_{2 \to 1} = 2 \times 10^7 \, \text{s}^{-1} \) and \( A_{1 \to 0} = 10^8 \, \text{s}^{-1} \). If \( 10^{14} \, \text{atoms/cm}^3 \) are excited from level 0 to level 2 and a steady state population in level 2 is achieved, then the steady state population at level 1 will be \( x \times 10^{13} \, \text{cm}^{-3} \). The value of \( x \) (in integer) is \(\underline{\hspace{2cm}}\).
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In steady state conditions for atomic systems, the excitation and de-excitation rates must be balanced.
In a steady state, the rate of excitation from level 0 to level 2 equals the rate of de-excitation from level 2 to level 1. This can be expressed as:
\[
A_{2 \to 1} \cdot N_2 = A_{1 \to 0} \cdot N_1,
\]
where:
- \( N_1 \) and \( N_2 \) are the populations of levels 1 and 2, respectively.
From the problem statement, the total excitation rate is given as \( 10^{14} \, \text{atoms/cm}^3 \), so the population at level 2 is:
\[
N_2 = 10^{14} \, \text{atoms/cm}^3.
\]
Using the steady state condition:
\[
2 \times 10^7 \cdot 10^{14} = 10^8 \cdot N_1.
\]
Solving for \( N_1 \):
\[
N_1 = \frac{2 \times 10^7 \cdot 10^{14}}{10^8} = 2 \times 10^{13} \, \text{atoms/cm}^3.
\]
Thus, the value of \( x \) is 2.
