A photon is incident upon a hydrogen atom ejects an electron with a kinetic energy 10.7 eV. If the ejected electron was in the first excited state. Energy of photon incident nearly will be :

14.1 eV

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20.9 eV

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24.3 eV

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10.2 eV

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Solution

The correct option is A 14.1 eV
Total energy of electron in first excited state $T . E = - 3.4 e V$
Kinetic energy of electron $K . E = 10.7 e V$
Energy of photon $E = K . E + | T . E | = 10.7 + 3.4 = 14.1 e V$

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A photon is incident upon a hydrogen atom ejects an electron with a kinetic energy 10.7 eV. If the ejected electron was in the first excited state. Energy of photon incident nearly will be :

The correct option is A 14.1 eVTotal energy of electron in first excited state T.E=−3.4 eVKinetic energy of electron K.E=10.7 eVEnergy of photon E=K.E+|T.E|=10.7+3.4=14.1 eV

The correct option is A 14.1 eV
Total energy of electron in first excited state $T . E = - 3.4 e V$
Kinetic energy of electron $K . E = 10.7 e V$
Energy of photon $E = K . E + | T . E | = 10.7 + 3.4 = 14.1 e V$