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Chapter 7: Problem 176

When the excited electron in a hydrogen atom falls from \(n=5\) to \(n=2,\) a photon of blue light is emitted. If an excited electron in He \(^{+}\) falls from \(n=4,\) to which energy level must it fall so that a similar blue light (as with the hydrogen) is emitted? Prove it. (See Exercise \(174 . )\)

Short Answer

Expert verified
The excited electron in the helium ion must fall to the energy level approximately equal to \(+\sqrt{45}\) (≈ 6.71) to emit a similar blue light as when an electron falls from n=5 to n=2 in the hydrogen atom. This is derived using the Rydberg formula for the energy difference, setting up the equation for the ratio of energy differences, and solving for n.

Step by step solution

01

Calculate the energy difference for hydrogen emission

First, we will find the energy difference when an electron falls from n=5 to n=2 in a hydrogen atom. To do this, we'll use the Rydberg formula for the energy difference: \(ΔE_n = -13.6 eV * \frac{Z^2}{n^2}\) Where: - \(ΔE_n\) is the energy difference - \(13.6 eV\) is the Rydberg constant for hydrogen (approximated value) - Z is the atomic number, which is 1 for hydrogen - n is the energy level to which the electron falls. The energy difference for hydrogen atom is: \(ΔE_{H} = -13.6 eV * \frac{1^2}{5^2} + 13.6 eV * \frac{1^2}{2^2}\)
02

Calculate the energy difference for the helium ion emission

Now, we will calculate the energy difference for helium ion emission using the Rydberg formula, keeping in mind that Z=2 for Helium: \(ΔE_{He} = -13.6 eV * \frac{2^2}{4^2} + 13.6 eV * \frac{2^2}{n^2}\) Where n is the energy level to which the electron falls in the helium ion.
03

Set up the ratio of the energy differences and solve for n

We know that the ratio of the energy differences for hydrogen and helium ion should be equal. So, we set up the equation and solve for n: \(\frac{ΔE_H}{ΔE_{He}} = 1\) Substitute the energy difference expressions from steps 1 and 2: \(\frac{-13.6 * \frac{1^2}{5^2} + 13.6 * \frac{1^2}{2^2}}{-13.6 * \frac{2^2}{4^2} + 13.6 * \frac{2^2}{n^2}} = 1\) Now, we can cancel out \(13.6 eV\) on both sides and solve for n: \(\frac{\frac{1^2}{5^2} - \frac{1^2}{2^2}}{\frac{2^2}{4^2} - \frac{2^2}{n^2}} = 1\)
04

Solve the equation for n

Next, we need to solve the equation obtained in step 3 for n: \(n^2 = 45\) Square root of both sides: \(n = \pm\sqrt{45}\) Since only the positive value of n is meaningful in this context, we consider n = \(+\sqrt{45}\). So the excited electron in the helium ion must fall to the energy level approximately equal to \(+\sqrt{45}\) (≈ 6.71) to emit a similar blue light as when an electron falls from n=5 to n=2 in the hydrogen atom.

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