i) Find the Kinetic energy of the electron when the hydrogen atom is in the first excited state.
ii) Find the kinetic energy, potential energy and the total energy of electron in first and second orbit of hydrogen atom, if potential energy in first orbit is taken to be zero.
Doubt by Aishwarya
Solution :
In ground state (n=1) of hydrogen atom
E1=-13.6 eV (Given)
E1=-13.6 eV (Given)
i) In first excited state, the electron jump to second orbit i.e. n=2
Total Energy (TE) of the electron in second orbit of the hydrogen atom is given by
Total Energy (TE) of the electron in second orbit of the hydrogen atom is given by
E2 = -13.6/n²
= -13.6/(2)²
= -13.6/4
= -13.6/(2)²
= -13.6/4
= -3.4 eV
TE = -3.4 eV
KE = -TE
KE = -(-3.4)
KE = 3.4 eV
KE = -(-3.4)
KE = 3.4 eV
ii)
In first orbit of Hydrogen atom
n=1
TE=E1
=-13.6/(1)²
=-13.6/(1)²
PE=2TE
=2(-13.6)
= -27.2 eV
KE=-TE
KE=-(-13.6)
KE=13.6 eV
KE=-(-13.6)
KE=13.6 eV
In second orbit of hydrogen atom
n=2
TE=E2
=-13.6/(2)²
=-13.6/(2)²
=-13.6/4
=-3.4 eV
PE=2TE
=2(-3.4)
= -6.8 eV
KE=-TE
KE=-(-3.4)
KE=3.4 eV
KE=-(-3.4)
KE=3.4 eV
When the potential energy in the first orbit is taken to be zero then the potential energies of each orbit would be increased by 27.2 eV.
For first orbit (n=1)
PE = -27.2 eV + 27.2 eV= 0 eV
KE = 13.6 eV
TE = PE+KE = 0 + 13.6 = 13.6 eV
For second orbit (n=2)
PE = -6.8 eV + 27.2 eV = 20.4 eV
KE = 3.4 eV
TE = PE+KE = 20.4 eV + 3.4 eV = 23.8 eV
