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In an excited state of hydrogen like ato...

In an excited state of hydrogen like atom an electron has total energy of `-3.4 eV`. If the kinetic energy of the electron is E and its de-Broglie wavelength is `lambda`, then

A

`E = 6.8 eV, lambda = 6.6 xx 10^(10)`

B

`E = 3.4 eV, lambda = 6.6 xx 10^(-10) m`

C

`E = 3.4 eV, lambda = 6.6 xx 10^(-11) m`

D

`E = 6.8 eV, lambda = 6.6 xx 10^(-11) m`

Text Solution

Verified by Experts

The correct Answer is:
B

`P.E. = -2(K.E.)`
`T.E. = (P.E.) + (K.E.)`
`T.E. = - 2(K.E.) + (K.E.)`
`T.E. = - (K.E.) - T.E. = K.E. , K.E. = 3.4 eV`
`lambda = 6.6 xx 10^(-10) m`
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Knowledge Check

  • The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state ?

    A
    `K = - 13.6 eV, U = -27.2 eV`
    B
    `K = 13.6 eV, U = -27.2 eV`
    C
    `K = - 13.6 eV, U = 27.2 eV`
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  • The ground state energy of hydrogen atom is - 13.6 eV. What is the potential energy of the electron in this state ?

    A
    0eV
    B
    `-27.2 eV`
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    `1 eV`
    D
    `2e V`
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