Atomic Structure Question 393
Question: When electronic transition occurs from higher energy state to lower energy state with energy difference equal to $ \Delta E $ electron volts, the wavelength of the line emitted is approximately equal to
Options:
A) $ \frac{12395}{\Delta E}\times {10^{-10}}m $
B) $ \frac{12395}{\Delta E}\times 10^{10}m $
C) $ \frac{12395}{\Delta E}\times {10^{-10}}cm $
D) $ \frac{12395}{\Delta E}\times 10^{10}cm $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \lambda =\frac{hc}{\Delta E} $
$ =\frac{6.62\times {10^{-34}}Js\times 3\times 10^{8}m{s^{-1}}}{E\times 1.602\times {10^{-19}}J} $
$ =\frac{12395}{E}\times {10^{-10}}m $
