When a hydrogen atom is excited from ground state to first excited state, then a)its kinetic energy increased by 20 eV b)its kinetic energy decreased by 10.2 eV c)its potential energy increased by 20.4 eV d)its angular momentum increased by 1.05×10^-34Js

Options

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`