(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.For the first excited st

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.For the first excited state,the principal quantum number is $n = 2$.Substituting $n = 2$ into the formula,we get $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.Therefore,the energy of the first excited state of hydrogen is $-3.4 \ eV$.

Class 11 Chemistry Structure of Atom Atomic models and Planck's quantum theory

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Which of the following is the energy of a possible excited state of hydrogen? $........... \ eV$

JEE MAIN 2015 All JEE MAIN

A
$-3.4$
B
$+6.8$
C
$+13.6$
D
$-6.8$

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Structure of Atom

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Atomic models and Planck's quantum theory

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Which of the following is the energy of a possible excited state of hydrogen? $........... \ eV$

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