An electron that has an energy of about 6 eV moves between rigid walls 1.00 nm apart. Find a) the quantum number n for the state of energy that the electron occupies, and b) the precise energy of the electron.
Added by Shawn G.
Step 1
We can rearrange this formula to solve for n: n = sqrt(8mLE / h²) Substituting the given values and constants (m = 9.11 x 10^-31 kg, L = 1.00 x 10^-9 m, E = 6 eV = 6 x 1.6 x 10^-19 J, h = 6.63 x 10^-34 J.s), we get: n = sqrt(8 * 9.11 x 10^-31 kg * 1.00 x 10^-9 Show more…
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