Question

Na atom remains in lowest excited state energy for a time 1.6 x 10${}^{-8}$ s before it makes a transition to a ground state emitting a photon of wavelength 589 nm. The wave length spread corresponding to this line will be

• A. 10${}^{-4}$nm
• B. 10${}^{-5}$nm
• C. 10${}^{-6}$nm
• D. 10${}^{-2}$nm

Answer 1

Answer: (A)

10${}^{-4}$nm

Uncertainty principle:
$\Delta E\Delta t=\frac{h}{4\pi }$
$hc\frac{\Delta \lambda }{{\lambda }^2}=\frac{h}{4\pi }$

Put $\lambda =589nm$
$\Delta \lambda ={10}^{-4}nm$