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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

  • An electron is in an excited state in a hydrogen like atom. It has a total energy of -3.4eV . The kinetic energy of the electron is E and its de Broglie wavelength is lamda

    A
    `E=6.8eV,lamda~6.6xx10^(-10)m`
    B
    `E=3.4eV,lamda~6.6xx10^(-10)m`
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    `E=6.8eV,lamda~6.6xx10^(-11)m`

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    0.168 eV
    B
    16.6 eV
    C
    1.68 eV
    D
    2.5 eV

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    B
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