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Find the energy for a hydrogen atom in the stationary state \(n=4.\)

Short Answer

Expert verified
Answer: The energy of a hydrogen atom in the stationary state with principal quantum number n = 4 is approximately -0.85 electron-volts.

Step by step solution

01

Recall the energy level formula for a hydrogen atom

The energy level formula for a hydrogen atom is given by: $$E_n = -\frac{13.6\,\text{eV}}{n^2}$$ where \(E_n\) is the energy of the stationary state with quantume number n, and eV represents electron-volt.
02

Plug the given value of n into the energy level formula

We are given that the principal quantum number for the stationary state is n = 4. Plug this value into the energy level formula: $$E_4 = -\frac{13.6\,\text{eV}}{(4)^2}$$
03

Calculate the energy of the hydrogen atom

Now we just have to simplify the expression: $$E_4 = -\frac{13.6\,\text{eV}}{16}$$ $$E_4 = -0.85\,\text{eV}$$ The energy of a hydrogen atom in the stationary state with principal quantum number n = 4 is approximately -0.85 electron-volts.

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