2.5. THEBOHRATOM 31
Bohrto guessthat,ifelectron angularmomentumcan onlychangebyunitsof ̄h,
thenthetotalangularmomentumoftheelectronintheHydrogenatomshouldbean
integermultipleofthatamount. Thisisthecondition(2.31). Letsseehowitleads
toapredictionforatomicspectra.
Theelectronisassumedtobemovingaroundthenucleusinacircularorbit.Now,
foranycircularorbit,thecentripetalforcemustequaltheattractiveforce,i.e.
p^2
mr
=
e^2
r^2
(2.35)
However,Bohr’squantizationcondition(2.31)implies
pn=
n ̄h
r
(2.36)
wherethesubscriptindicatesthat eachmomentumisassociatedwithaparticular
integern.Insertingthisexpressioninto(2.35)andsolvingforr,wefind
rn=
n^2 h ̄^2
me^2
(2.37)
Thetotalenergyoftheelectron,inanorbitofradiusrn,istherefore
En =
p^2 n
2 m
−
e^2
rn
=
n^2 h ̄^2
2 mrn^2
−
e^2
rn
= −
(
me^4
2 ̄h^2
)
1
n^2
(2.38)
Thetotalenergyisnegativebecausetheelectronisinaboundstate;energymustbe
addedtobringtheelectronenergytozero(i.e. afreeelectronatrest).
Bohr’sideawasthataHydrogenatomemitsaphotonwhentheelectronjumps
froman energy stateEm, toa lower energystate En. Thephoton energyis the
differenceofthesetwoenergies
Ephoton = Em−En
hf =
(
me^4
2 ̄h^2
)(
1
n^2
−
1
m^2
)
h
c
λ
=
(
me^4
2 ̄h^2
)(
1
n^2
−
1
m^2
)
1
λ
=
(
me^4
2 chh ̄^2
)(
1
n^2
−
1
m^2
)
(2.39)