Atomic Structure Question 236
Question: Based on the equation: $ \Delta E=-2.0\times {10^{-18}}J( \frac{1}{n_2^{2}}-\frac{1}{n_1^{2}} ) $ the wavelength of the light that must be absorbed to excite hydrogen electron from level n = 1 to level $ n=2 $ will be: ( $ h=6.625\times {10^{-34}}Js $ , $ C=3\times 10^{8}m{s^{-1}} $ )
Options:
A) $ 1.325\times {10^{-7}}m $
B) $ 1.325\times {10^{-10}}m $
C) $ 2.650\times {10^{-7}}m $
D) $ 5.300\times {10^{-10}}m $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \Delta E=-2.0\times {10^{-18}}\times ( \frac{1}{2^{2}}-\frac{1}{1^{2}} ) $
$ =-2.0\times {10^{-18}}\times \frac{-3}{4} $
$ =1.5\times {10^{-18}} $
$ \Delta E=\frac{hc}{\lambda } $
$ \lambda =\frac{hc}{\Delta E}=\frac{6.6\times {10^{-34}}\times 3\times 10^{8}}{1.5\times {10^{-18}}} $
$ =1.325\times {10^{-7}}m $
BETA
