Atomic Structure Question 378

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 $

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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 $



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