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Chapter 19: Problem 68

In hydrogen atom, if potential energy of electron in ground state is assumed to be zero, then its energy in first excited state is equal to (A) \(-3.4 \mathrm{eV}\) (B) \(23.8 \mathrm{eV}\) (C) \(17.2 \mathrm{eV}\) (D) \(-6.8 \mathrm{eV}\)

Short Answer

Expert verified
Using the energy level formula for the hydrogen atom, \(E_{n} = -\frac{13.6\,\text{eV}}{n^2}\), we find the energy of the electron in the first excited state (n=2) to be \(E_{2} = -\frac{13.6\,\text{eV}}{2^2} = -3.4\,\text{eV}\). Therefore, the correct answer is (A) \(-3.4\,\text{eV}\).

Step by step solution

01

Recall the energy level formula for the hydrogen atom

We can find the energy of an electron in any energy level (n) in a hydrogen atom using the following formula: \[E_{n} = -\frac{13.6\,\text{eV}}{n^2}\] where \(E_{n}\) is the energy of the electron in the nth energy level and n is the principal quantum number.
02

Calculate the energy of the electron in the first excited state

In the first excited state, the electron's principal quantum number is n = 2. Let's plug this value into the energy level formula. \[E_{2} = -\frac{13.6\,\text{eV}}{2^2}\]
03

Simplify the expression

Now we simply need to calculate the energy by simplifying the expression: \[E_{2} = -\frac{13.6\,\text{eV}}{4}\] \[E_{2} = -3.4\,\text{eV}\]
04

Choose the correct option

The calculated energy of the electron in the first excited state is \(-3.4\,\text{eV}\). Thus, the correct answer is: (A) \(-3.4\,\text{eV}\)

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