Question

energy of electron in the first excited state in hydrogen atom is -3.4 electron volt find kinetic energy and potential energy of electron in the ground state​

Answer 1

KE: 3.4ev

PE:6.8ev

Explanation:

|KE|= -TE

-(-3.4ev) =3.4ev

PE=2×|TE|=6.8ev

Hope u understand..

Answer 2

K.E = 3.4 eV and P.E = - 6.8 eV .

Given :

Energy of electron in the first exited state of hydrogen atom is , E = -3.4 eV .

Now, we need to find Kinetic and Potential energy of the electron.

For , this we should know the relation of total energy between kinetic energy and potential energy .

K.E = - ( E ) and P.E = 2E  

[  NOTE : We should not apply these formulas in classical physics . These formula comes after some long derivation , which are not in scope ]

Therefore , K.E = 3.4 eV and P.E = - 6.8 eV .

Learn More :

Energy of electrons .

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