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Question

The ground state energy of hydrogen atom is 13.6 eV. When its electron is in the first excited state, its excitation energy is:


A
10.2 eV
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B
zero
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C
3.4 eV
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D
6.8 eV
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Solution

The correct option is B 10.2 eV
The energy of an electron in Bohr’s orbit of hydrogen atom is given by the expression:

$$E_n=-13.6 \dfrac{Z^2}{n^2}$$; [where n = energy state of electron]

For hydrogen Z = 1

$$E_1= -13.6 eV$$ (given)

For the first excited state, n = 2,

Therefore, $$E_2=\dfrac{-13.6}{2^2} eV$$ 

$$E_2= -3.4\ eV$$

Now, $$\Delta{E}=E_2-E_1$$ 

$$= -3.4 - (-13.6) = 10.2 eV$$

Chemistry

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