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Calculate the enegry of an electron in t...

Calculate the enegry of an electron in the second Bohr orbit of an `H` atom.

Text Solution

Verified by Experts

`E_n = -(13.6)/(n^2) eV`
Energy of first excited state (n = 2),
`E_2 = -(13.6)/(2^2) eV = -3.4 eV`.
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Knowledge Check

  • The energy of an electron in second Bohr orbit of hydrogen atom is :

    A
    `- 5. 44 xx 10^(-19) eV`
    B
    `- 5.44 xx 10 ^(-19) cal `
    C
    `-5.44 xx 1 ^(-19) KJ`
    D
    ` - 5.44 xx 10 ^(-19) J`

  • The energy of an electron in the nth Bohr orbit of hydrogen atom is

    A
    `-(13.6)/(n^(4))`eV
    B
    `-(13.6)/(n^(3))`eV
    C
    `-(13.6)/(n^(2)) ` eV
    D
    `-(13.6)/(n) ` eV