Calculating Frequency of Absorbed Electromagnetic Radiation in Bohr Atom

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The energy of an electron in a Bohr atom is calculated using the formula E_n = -2.179 x 10^-18/n^2 J. To find the frequency of electromagnetic radiation absorbed when an electron transitions from level n = 4 to n = 9, the energy difference between these levels must first be calculated. This difference is then related to frequency using the equation ΔE = hv, where h is Planck's constant. Users confirmed that the calculation steps involving plugging in values, subtracting energies, and dividing by Planck's constant were correct. The discussion concludes with a successful calculation of the frequency, indicating effective collaboration and problem-solving.
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The energy (in joules) of an electron energy level in the Bohr atom is given by the expression: E_{n}= -2.179 x 10^-18/n^2 Jwhere n is the principal quantum number for the energy level. What is the frequency in Hz of the electromagnetic radiation absorbed when an electron is raised up from level with n = 4 to that with n = 9?

I'm unsure of the the formula to use, or even the first step. Any ideas??
 
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First use your formula to calculate the energy difference between the two levels. Second, relate that to frequency. Look in your book for the formula giving energy of a photon (hint: it involves Planck's constant). You can show us what you get if you still have questions.
 
use the equation \Delta E = R_{H}^{*} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}) = hv
 
Ok, so I plugged both of the numbers (4 and 9) into the equation. Then I subtracted answer #1 from answer #2. Then I divided this number by h (6.626E-34). Am I doing this correctly?
 
Yes, exactly.
 
this is great!, I could learn something here...
 
Great I got the answer, thanks for all the help!
 
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