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

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A. its kinetic energy increases by 10.2 eV
B. its kinetic energy decreases by 10.2 eV
C. its potential energy increases by 20.4 eV
D. its angular momentum increases by `1.05xx10^(-34)J-s`

Explanation

In ground state n=1 and for first excited state n=2
`KE=1/(4piepsilon_(0))(e^(2))/(2r)(z=1)=(14.4xx10^(-10))/(2r) eV ( :' r=0.53 n^(2) A^(0) (z=1))`
`(KE)_(1)=(14.4xx10^(-10))/(2xx0.53xx10^(-10)) eV=13.58 eV` and `(KE)_(2)=(14.4xx10^(-10))/(2xx0.53xx10^(-10)xx4) ev=3.39 ev`
`:.` Ke decreasess by 10.2 ev
`:.` PE increases by =Excitation energy + Loss in kinetic energy =10.2+10.2=20.4 ev
Now angular momentum, `L=mvr=(nh)/(2pi)`
`implies L_(2)-L_(1)=h/(2pi)=(6.6xx10^(-34))/6.28=1.05xx10^(-34) J-sec`