If potential energy of an electron in a hydrogen atom in first excited state is taken to be zero, kinetic energy (in eV) of an electron in ground state will be

Question

If potential energy of an electron in a hydrogen atom in first excited state is taken to be zero, kinetic energy (in eV) of an electron in ground state will be (A) 13.6 eV (B) 10.2 eV (C) 3.4 eV (D) 5.1 eV

Tardigrade Admin · Accepted Answer

If zero of potential energy is changed,

K E

does not change and continues to be

+ 13.6, e V

.
When

2^{nd}

shell is taken as reference,
ground state

PE

= - 13.6 e V \times 2 + 6.8 e V

= - 20.4 e V

K E = \frac{- PE}{2}

= 10.2 e V

Q. If potential energy of an electron in a hydrogen atom in first excited state is taken to be zero, kinetic energy (in eV) of an electron in ground state will be

13.6 eV

10.2 eV

3.4 eV

5.1 eV

If zero of potential energy is changed, KE does not change and continues to be +13.6,eV. When 2nd shell is taken as reference, ground state PE =−13.6eV×2+6.8eV =−20.4eV KE=2−PE​ =10.2eV

If zero of potential energy is changed,
$K E$ does not change and continues to be $+ 13.6, e V$ .
When $2^{nd}$ shell is taken as reference,
ground state $PE$
$= - 13.6 e V \times 2 + 6.8 e V$
$= - 20.4 e V$
$K E = \frac{- PE}{2}$
$= 10.2 e V$