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Chapter 38: Problem 59

What is the energy of the orbiting electron in a hydrogen atom with a quantum number of \(45 ?\)

Short Answer

Expert verified
Answer: The energy of the orbiting electron in a hydrogen atom with a quantum number of 45 is approximately -0.0067 eV.

Step by step solution

01

Write down the given information and formula

The quantum number of the hydrogen atom is given: \(n=45\). The formula for the energy levels of the hydrogen atom is: \(E_n = -\frac{13.6 \text{eV}}{n^2}\)
02

Substitute the given quantum number in the formula

We need to find the energy of the orbiting electron in the hydrogen atom. Substitute the given quantum number \(n=45\) in the energy formula: \(E_{45} = -\frac{13.6 \text{eV}}{45^2}\)
03

Calculate the energy of the electron

Now, calculate the energy of the electron in the hydrogen atom with the given quantum number: \(E_{45} = -\frac{13.6 \text{eV}}{45^2} = -\frac{13.6 \text{eV}}{2025} = -0.006716049 \text{eV}\) The energy of the orbiting electron in a hydrogen atom with a quantum number of \(45\) is approximately \(-0.0067 \text{eV}\).

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Most popular questions from this chapter

A low-power laser has a power of \(0.50 \mathrm{~mW}\) and a beam diameter of \(3.0 \mathrm{~mm}\). a) Calculate the average light intensity of the laser beam, and b) compare it to the intensity of a 100 -W light bulb producing light viewed from \(2.0 \mathrm{~m}\).

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