Structure of Atom - Result Question 16

####16. In hydrogen atom, energy of first excited state is 3.4eV. Find out KE of the same orbit of Hydrogen atom

[2002]

(a) +3.4eV

(b) +6.8eV

(c) 13.6eV

(d) +13.6eV

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Solution:

  1. (a) Suppose the nucleus of hydrogen atom have charge of one proton +e. The electron revolves in an orbit of radius r around it. Therefore the centripetal force is supplied by electrostatic force of attraction between the electron and nucleus i.e.

mv2r=Ze2r2

or mv2=Ze2r

or 12mv2=12Ze2r=K.E

now total energy (En)=K.E+ P.E

in first excited state

E=12mv2+[Ze2r]

=+12Ze2rZe2r

3.4eV=12Ze2r

K.E=12Ze2r=+3.4eV

For Bohr orbit, kinetic energy = Total energy and Potential energy =2× Total energy.



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