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  • When a hydrogen atom is excited from ground state to first excited state then which is incorrect?Option: 1 its kinetic energy increases by <img alt="

When a hydrogen atom is excited from ground state to first excited state then which is incorrect?

Option: 1

its kinetic energy increases by 10.2\, \, eV.


Option: 2

its kinetic energy decreases by 10 .2 \, eV.


Option: 3

its potential energy increases by 20.4\, eV.


Option: 4

 its angular momentum increases by 1.05 \times 10^{-34} \mathrm{~J}-\mathrm{s}.


Answers (1)

best_answer

We have total energy

                =-\frac{13.6}{n^2} \mathrm{eV}

            Also

                 \text { T.E. }=-\mathrm{K} . \mathrm{E} .

                  =\frac{1}{2} \text { P.E. }

                \text { For } n=1

               \text { T.E. }=-13.6 \mathrm{eV}, \text { K.E. }=+13.6 \mathrm{eV}, \text { P.E. }=-27.2 \mathrm{eV}

                \text { For } n=2

                \text { T.E. }=-3.4 \mathrm{eV}, \mathrm{K} . E .=+3.4 \mathrm{eV}, \mathrm{P} . \mathrm{E} .=-6.8 \mathrm{eV}

                 \therefore \quad \text { PE increases by }=20.4 \mathrm{ev}

                  Now Angular momentum ;

                                          \mathrm{L}=\mathrm{mvr}=\frac{n h}{2 \pi}

               \mathrm{L}_2-\mathrm{L}_1=\frac{h}{2 \pi}=\frac{6.6 \times 10^{-34}}{6.28}=1.05 \times 10^{-34} \mathrm{~J}-\mathrm{sec} .

Posted by

jitender.kumar

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