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Chapter 27: Problem 64

How much energy is required to remove an electron from a hydrogen atom in the \(n=4\) state?

Short Answer

Expert verified
Answer: The ionization energy required is 0.85 eV.

Step by step solution

01

Recall the Rydberg formula

The Rydberg formula for the energy levels of a hydrogen atom is given by: \(E_n = -\dfrac{13.6 eV}{n^2}\) where \(E_n\) is the energy of the nth level, and n is the principal quantum number.
02

Calculate the energy of the n=4 state

To find the energy of the n=4 state, we substitute n=4 into the Rydberg formula: \(E_4 = -\dfrac{13.6 eV}{4^2} = -\dfrac{13.6 eV}{16} = -0.85 eV\)
03

Calculate the energy difference between the n=4 and n=∞ states

To remove an electron from a hydrogen atom, we need to ionize it, which means bringing the electron from its current energy level (n=4) to the infinite energy level (n=∞). The energy of the n=∞ state is 0 eV since, at infinity, the electron is completely free of the nucleus. So the energy difference between the n=4 and n=∞ states is: \(\Delta E_{ionization} = E_{\infty} - E_4 = 0 - (- 0.85 eV)\)
04

Calculate the ionization energy

Now we can calculate the ionization energy as follows: \(\Delta E_{ionization} = 0.85 eV\) This means that 0.85 electron volts of energy is required to remove an electron from a hydrogen atom in the n=4 state.

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Most popular questions from this chapter

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