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A difference of 2.3 eV separates two ene...

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level ?

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It is given here `E_(n_(1)) - E_(n_(2)) = 2.3 eV = 2.3 xx 1.6 xx 10^(-19)` J
`therefore ` radiation frequecy ` v= (E_(n_(1)) - E_(n_(2)))/(h) = (2.3 xx 1.6 xx 10^(-19))/(6.63 xx 10^(-34)) = 5.6 xx 10^(14)Hz`
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Knowledge Check

  • Two energy levels of an electron in an atom are separated by 2.3 eV. The frequency of radiation emitted when the electrons go from higher to lower level is

    A
    `6.95xx10^(14)Hz`
    B
    `3.68xx10^(15)Hz`
    C
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    D
    `9.11xx10^(15)Hz`

  • The wavelength of the spectral line when the electron is the hydrogen atom undergoes a transition from the energy level 4 to energy level 2 is.

    A
    486 nm
    B
    486 m
    C
    `486 Å`
    D
    486 cm

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