Nuclear Physics And Radioactivity Question 27
Question: A sodium atom is in one of the states labeled ‘Lowest excited levels’. It remains in that state for an average time of 10?8 sec, before it makes a transition back to a ground state. What is the uncertainty in energy of that excited state
Options:
A) 6.56 ´ 10?8 eV
B) 2 ´ 10?8 eV
C) 10?8 eV
D) 8 ´ 10?8 eV
Show Answer
Answer:
Correct Answer: A
Solution:
The average time that the atom spends in this excited state is equal to Dt, so by using $ \Delta E.,\Delta t=\frac{h}{2\pi } $
Þ Uncertainty in energy $ =\frac{h/2\pi }{\Delta t} $ $ =\frac{6.6\times {{10}^{-34}}}{2\times 3.14\times {{10}^{-8}}}=1.05\times {{10}^{-26}}J=6.56\times {{10}^{-8}}eV $
BETA
