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Question

The energy of an electron in the Bohr's first orbit of $$H$$-atom is $$-13.6eV$$. The possible energy value (s) of the excited state (s) for electrons in Bohr's orbits of hydrogen is (are):


A
3.4eV
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B
4.2eV
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C
6.8eV
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D
+6.8eV
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Solution

The correct option is A $$-3.4eV$$
As we know,

$$E = -13.6\times Z^2/n^2$$ 

So, $$E_1$$ for $$H=-13.6eV$$

$$E_2=\dfrac{-13.6}{4}=-3.4eV$$;

$$E_3=\dfrac{-13.6}{9}=-1.51;$$
$$E_4=\dfrac{-13.6}{16}=-0.85$$ and so on.

Hence, the correct option is $$A$$

Chemistry

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