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Question

The energies of the 3 lowest states of an atom are E 0 = −14 eV, E 1  = −9 eV and E 2  = −7 eV. The Einstein coefficients are A 10  = 3 × 10 8  s −1 , A 20  = 1.2 × 10 8  s −1  and A 21  = 8 × 10 7  s −1 . If a large number of atoms are in the energy level E 2 , the mean radiative lifetime of this excited state is

The correct answer is 0.5 × 10 −8 s

Mean Radiative Lifetime Explained

The mean radiative lifetime of an excited atomic state represents the average time an atom spends in that state before it transitions to a lower energy state by emitting a photon. This lifetime is determined by the total probability rate of spontaneous emission from that excited state to all possible lower energy states.

Einstein Coefficients and Transition Rates

Einstein coefficients, specifically the Einstein A coefficients ($A_{ij}$), quantify the spontaneous emission probability rate for a transition from a higher energy level $i$ to a lower energy level $j$. If an atom is in an excited state, say state 2, it can transition to any lower energy state, say state 1 or state 0, by spontaneous emission. The rate of transition from state 2 to state 1 is $A_{21}$, and the rate of transition from state 2 to state 0 is $A_{20}$.

The total transition probability rate from state 2 is the sum of the rates of all possible downward transitions:

$$ \text{Total transition rate from state 2} = A_{21} + A_{20} $$

Calculating Mean Radiative Lifetime

The mean radiative lifetime, denoted by $\tau$, of an excited state is the reciprocal of the total transition probability rate from that state. For state 2, the mean radiative lifetime $\tau_2$ is given by:

$$ \tau_2 = \frac{1}{\text{Total transition rate from state 2}} = \frac{1}{A_{21} + A_{20}} $$

Applying Given Values

We are given the following Einstein coefficients for transitions from state 2:

  • $A_{20} = 1.2 \times 10^8 \text{ s}^{-1}$ (Transition from state 2 to state 0)
  • $A_{21} = 8 \times 10^7 \text{ s}^{-1}$ (Transition from state 2 to state 1)

Note that the energies of the states are given as $E_0 = -14$ eV, $E_1 = -9$ eV, and $E_2 = -7$ eV. Since $E_2 > E_1 > E_0$, transitions from E2 can occur to both E1 and E0.

Now, we can calculate the total transition rate from state 2:

$$ \text{Total rate} = A_{20} + A_{21} $$
$$ \text{Total rate} = (1.2 \times 10^8 \text{ s}^{-1}) + (8 \times 10^7 \text{ s}^{-1}) $$

Let's express both coefficients with the same power of 10:

$$ 8 \times 10^7 \text{ s}^{-1} = 0.8 \times 10^8 \text{ s}^{-1} $$
$$ \text{Total rate} = (1.2 \times 10^8 \text{ s}^{-1}) + (0.8 \times 10^8 \text{ s}^{-1}) $$
$$ \text{Total rate} = (1.2 + 0.8) \times 10^8 \text{ s}^{-1} $$
$$ \text{Total rate} = 2.0 \times 10^8 \text{ s}^{-1} $$

Now, we calculate the mean radiative lifetime $\tau_2$:

$$ \tau_2 = \frac{1}{\text{Total rate}} = \frac{1}{2.0 \times 10^8 \text{ s}^{-1}} $$
$$ \tau_2 = \frac{1}{2.0} \times 10^{-8} \text{ s} $$
$$ \tau_2 = 0.5 \times 10^{-8} \text{ s} $$

Conclusion

The calculated mean radiative lifetime of the excited state E2 is $0.5 \times 10^{-8}$ s. This value represents the average time an atom will remain in the E2 state before spontaneously emitting a photon and transitioning to either the E1 or E0 state.

The final answer is $\boxed{0.5 \times 10^{-8} \text{ s}}$.

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Important Questions from Radioactivity

  1. The nuclei of 137 Cs decay by the emission of β - particles with a half life of 30.08 years. The activity (in units of disintegrations per second or Bq) of a 1 mg source of 137 Cs, prepared on January 1, 1980, as measured on January 1, 2021 is closest to

  2. The Q - value of the α - decay of 232 Th to the ground state of 228 Ra is 4082 keV. The maximum possible kinetic energy of the α - particle is closest to

  3. Radioactivity is the characteristic of which of the following?

  4. Particles which can be added to the nucleus of an atom without changing its chemical properties are

  5. A free neutron decays spontaneously into a proton, an electron and ___________ .

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