Audio visual focus module, atomic spectra, Q. 42

Megan Singer 3D · Postby **Megan Singer 3D** » Sat Oct 17, 2020 12:20 am

Can someone help me solve this problem from the audio visual focus modules?

42. An excited hydrogen atom emits light with a frequency of 1.14 x 10^14 Hz to reach the energy level n = 4. In what principle quantum level did the electron begin?

Options include n=5, n=6, n=4, and n=7.

Anh Trinh 1J · Postby **Anh Trinh 1J** » Sat Oct 17, 2020 12:55 am

For this question, I used the Rydberg equation and $E_{n}=\frac{-hR}{n^{2}}$ . First, you have to find the energy in energy level. Equation is $E_{n}=\frac{-hR}{n^{2}}$ . Plank Constant is $6.626\cdot 10^{-34}J$ and Rydberg constant is $3.29\cdot 10^{15} s^{-1}$ . After plugging in the constants and n, energy would then be $-1.36\cdot 10^{-19} J$ . Then you can use the Rydberg equation to find the initial energy level. The Rydberg equation is $V=R[\frac{1}{n^{2}_{1}} - \frac{1}{n^{2}_{2}}]$ . n=4 is the final energy level, so it would be $n_{1}$ . After all the calculations, $n_{2}$ would be 6.

There's another way you could also solve this problem:
$E=hv=6.626\cdot 10^{-34}Js(1.14\cdot 10^{14}m/s)= 7.55\cdot 10^{-20}J$

$E_{4}=\frac{-hR}{n^{2}}=\frac{-6.626\cdot 10^{-34}(3.29\cdot 10^{15}Hz)}{16}=-1.36\cdot 10^{-19}J$

$-1.36\cdot 10^{-19}J-E_{intial}=-7.55\cdot 10^{-20}J$

$E_{initial}=6.046\cdot 10^{-20}J$

$n^{2}=\frac{-hR}{6.046\cdot 10^{-20}J}=\sqrt{36}=6$

Madison Muggeo 3H · Postby **Madison Muggeo 3H** » Sat Oct 17, 2020 10:23 am

Thank you guys for this post! I was confused and the explanation really helped :)

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Audio visual focus module, atomic spectra, Q. 42

Audio visual focus module, atomic spectra, Q. 42

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