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The energy of the electron in the ground...

The energy of the electron in the ground state of H-atom is `-13.6 eV`. The energy of the first excited state will be

A

`-3.4 eV`

B

`-27.2 eV`

C

`-6.8 eV`

D

`-52.4 eV`

Text Solution

Verified by Experts

The correct Answer is:
A
`E_(n)=E_(1)/n^(2)`
Thus, energy of electron in the first excited state is given by,
`E_(2)=E_(1)/4=(-13.6)/4=-3.4 eV`
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Knowledge Check

  • The energy of the electron in the ground state of H-atom is -13.6 eV . Its energy in the second orbit is

    A
    `-13.6 eV`
    B
    `-3.4 eV`
    C
    `-1.51 eV`
    D
    `-0.85 eV`

  • The energy of an electron in the first Bohr orbit is -13 eV . The energy of Be^(3+) in the first excited state is :

    A
    `-30.6 eV`
    B
    `-40 .8 eV`
    C
    `-54.4` eV
    D
    `+40.8` eV

  • The energy of an electron in the first orbit of H- atom is -13.6eV . The possible energy values of the excited state for electrons in Bohr orbits of Li^(2+) ions is/are-

    A
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    B
    `-30.6eV`
    C
    `-13.6eV`
    D
    Both B & C are correct

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