Which of the following is the energy of a possible excited state of hydrogen?

+ 6.8 e V

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+ 13.6 e V

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- 6.8 e V

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- 3.4 e V

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Solution

The correct option is D $- 3.4 e V$
For 1st excited state, the value of n=2
By putting the value of n, we get the value of energy.

Thus, the energy of electron in first excited state of hydrogen atom $= - \frac{13.6}{n^{2}} = - \frac{13.6}{4} = - 3.4 e V$ .

So, the correct option is $D$

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Which of the following is the energy of a possible excited state of hydrogen?

The correct option is D −3.4eVFor 1st excited state, the value of n=2By putting the value of n, we get the value of energy.Thus, the energy of electron in first excited state of hydrogen atom =−13.6n2=−13.64=−3.4eV.So, the correct option is D

The correct option is D $- 3.4 e V$
For 1st excited state, the value of n=2
By putting the value of n, we get the value of energy.

Thus, the energy of electron in first excited state of hydrogen atom $= - \frac{13.6}{n^{2}} = - \frac{13.6}{4} = - 3.4 e V$ .

So, the correct option is $D$