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Chapter 18: Problem 2481

If \(13.6 \mathrm{eV}\) energy is required to ionise the hydrogen atom the energy required to remove the electron form \(n=2\) state is (A) Zero (B) \(10.2 \mathrm{eV}\) (C) \(6.8 \mathrm{eV}\) (D) \(3.4 \mathrm{eV}\)

Short Answer

Expert verified
The energy required to remove the electron from the n=2 state is \(10.2 \mathrm{eV}\).

Step by step solution

01

Calculate energy of n=2 state for hydrogen atom

We use the energy level formula for hydrogen atom: \[E_n = -\dfrac{13.6 \, \text{eV}}{n^2}\] Now substituting n=2 \[E_2 = -\dfrac{13.6 \, \text{eV}}{(2)^2} = -\dfrac{13.6 \, \text{eV}}{4} = -3.4\, \text{eV}\]
02

Calculate energy difference between n=1 and n=2 states

Since, the energy required to ionize the hydrogen atom (remove electron from n=1 state) is 13.6 eV, the energy of the ground state (n=1) is: \[E_1 = -13.6\, \text{eV}\] Now to find the energy required to remove the electron from the n=2 state, we calculate the energy difference between n=1 and n=2 states: \[\Delta E = E_2 - E_1 = (-3.4\, \text{eV}) - (-13.6\, \text{eV}) = 10.2\, \text{eV}\] So, the energy required to remove the electron from the n=2 state is 10.2 eV. The correct answer is (B) \(10.2 \mathrm{eV}\).

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