Question

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 \(\underline{\hspace{2cm}}\). 

Hint:

In steady state conditions for atomic systems, the excitation and de-excitation rates must be balanced.

Answer 1

Correct Answer: 2

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.