The total energy of the electron in the first excited state of hydrogen is $- 3.4 e V$ . What is the kinetic energy of the electron in this state?

+ 1.7 e V

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+ 3.4 e V

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+ 6.8 e V

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- 13.4 e V

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Solution

The correct option is B $+ 3.4 e V$
Total energy of an excited electron of hydrogen atom
=(-)Kinetic energy of that electron in that state.
or, T.E=P.E+K.E
and $K . E = \frac{K Z e^{2}}{2 r} a n d P . E = - \frac{K Z e^{2}}{r} o r T . E = (-) K . E$
Therefore, If
TE=(-)3.4eV
KE=3.4eV

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The total energy of the electron in the first excited state of hydrogen is −3.4 eV. What is the kinetic energy of the electron in this state?

The correct option is B +3.4 eVTotal energy of an excited electron of hydrogen atom=(-)Kinetic energy of that electron in that state.or, T.E=P.E+K.Eand K.E=KZe22randP.E=−KZe2rorT.E=(−)K.ETherefore, IfTE=(-)3.4eVKE=3.4eV

The total energy of the electron in the first excited state of hydrogen is $- 3.4 e V$ . What is the kinetic energy of the electron in this state?

The correct option is B $+ 3.4 e V$
Total energy of an excited electron of hydrogen atom
=(-)Kinetic energy of that electron in that state.
or, T.E=P.E+K.E
and $K . E = \frac{K Z e^{2}}{2 r} a n d P . E = - \frac{K Z e^{2}}{r} o r T . E = (-) K . E$
Therefore, If
TE=(-)3.4eV
KE=3.4eV