Answer to Question #87972 in Atomic and Nuclear Physics for Sridhar

Question

An electron in the n=2 state of hydrogen remains there on the average about 10^-8s before making a transition to n=1 state the uncertainty in the energy of an n=2 state is
Ans: 4.1×10^-7 ev

Answer 1

The uncertainty in energy is connected with the lifetime of the corresponding excited state via the following expression:

$$\Delta E \cdot \Delta t = \hbar$$

Hence:

$$\Delta E= \frac{\hbar}{\Delta t} = \frac{6.63 \cdot 10^{-34}}{10^{-8}} \, J \, = 6.63 \cdot 10^{-26} \, J$$

As long as

$$1 \, eV = 1.6 \cdot 10^{-19} \, J,$$

we obtain

$$\Delta E= \frac{6.63 \cdot 10^{-26}}{1.6 \cdot 10^{-19}} \approx 4.1 \cdot 10^{-7} \, eV$$

Answer: 4.1*10^-7 eV.