Question
A hydrogen atom is in its third excited state (n=4). Using Bohr theory of the atom calculate (a) the radius of the orbit, (b) the linear momentum of the electron, (c) the angular momentum of the electron (d) the kinetic energy (e) the potential energy, and (f) the total energy. (g) if the electron jumps to the ground state, what is the wavelength of the emitted photon?
Answers
r(n) = ε(o) •h^2•n^2/π•m•e^2 =
=5.3•10^-11•n^2 =5.3•10^-11•16 =
=8.48•10^-10 m.
v(n) = e^2/2• ε(o) •h•n=
=2.19•10^6/n =5.475•10^5 m/s.
p(n)=e•v(n) =1.6•10^-19•5.475•10^5 =
=8.77•106-14 kg•m/s.
L =m•v(n)•r(n) =n•h/2• π =4.22•10^-34 kg•m^2/s.
E(total) = -(1/4•π•ε(o)) •(e^2/2•r(n)) = -13.6/n^2 =
= - 0.85 eV = - 13.6•10^-19 J.
KE = - E(total) = = + 0.85 eV = + 13.6•10^-19 J.
PE = 2•E(total) = - 1.7 eV = - 2.72•10^-19 J.
1/λ = R•(1/1^2 – 1/n^2) =3•R/4,
λ =4/3•R=1.2•10^-7 m (Paschen series, infrared band)
=5.3•10^-11•n^2 =5.3•10^-11•16 =
=8.48•10^-10 m.
v(n) = e^2/2• ε(o) •h•n=
=2.19•10^6/n =5.475•10^5 m/s.
p(n)=e•v(n) =1.6•10^-19•5.475•10^5 =
=8.77•106-14 kg•m/s.
L =m•v(n)•r(n) =n•h/2• π =4.22•10^-34 kg•m^2/s.
E(total) = -(1/4•π•ε(o)) •(e^2/2•r(n)) = -13.6/n^2 =
= - 0.85 eV = - 13.6•10^-19 J.
KE = - E(total) = = + 0.85 eV = + 13.6•10^-19 J.
PE = 2•E(total) = - 1.7 eV = - 2.72•10^-19 J.
1/λ = R•(1/1^2 – 1/n^2) =3•R/4,
λ =4/3•R=1.2•10^-7 m (Paschen series, infrared band)