QMGreensite_merged

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)