1000 Solved Problems in Modern Physics

2.3 Solutions 103

νp=

ω

k

=

(

c^2 k^2 +

m^2 c^4

^2

) 1 / 2

/k

νg=

dω

dk

=kc^2

(

c^2 k^2 +

m^2 c^4

^2

)− 1 / 2

∴νpνg=c^2

2.3.2 Hydrogen Atom...................................

2.7 Apart from the principle quantum numbern, three other quantum numbers

are required to specify fully an atomic quantum state vizl, the orbital angular

quantum number,mlthe magnetic orbital angular quantum number, andms

the magnetic spin quantum number.

Forn= 1 ,l=0, if there is only one electron as in H-atom, then it will be

in 1sorbit. The total angular momentumJ =l± 1 /2, so thatJ = 1 /2. In

the spectroscopic notation,2s+^1 LJ, the ground state is therefore a^2 S 1 / 2 state.

Forn=2, the possible states are^2 S and^2 P. if there are two electrons as in

helium atom, both the electrons can go into theK-shell (n=1) only when

they have antiparallel spin direction (↑↓) on account of Pauli’s principle.

This is because if the spins were parallel, all the four quantum numbers would

be the same for both the electrons (n = 1 ,l = 0 ,ml = 0 ,ms=+ 1 /2).

Therefore in the ground stateS=0, and since both electrons are 1selectrons,

L=0. Thus the ground state is a S state (closed shell). A triplet state is not

given by this electron configuration. An excited state results when an electron

goes to a higher orbit. Then both electrons can have, in addition, the same

spin direction, that is we can haveS=1aswellasS=0 Excited triplet

and singlet spin states are possible (orthohelium and parahelium). The lowest

triplet has the electron configuration 1s 2 s,itisa^3 s 1 , state. It is a metastable

state. The corresponding singlet state is 2^1 S 0 , and lies somewhat higher.

Carbon has six electrons. The Pauli principle requires the ground state con-

figuration 1S^2 2S^2 2P^2. The superscripts indicate the number of electrons in a

given state.

2.8 A carbon atom has 6 electrons. If all these electrons are replaced byπ−

mesons then two differences would arise (i) Asπ−mesons are bosons (spin

0) Pauli’s principle does not operate so that all of them can be in the K-shell

(n=1) (ii) The total energy is enhanced because of the reduced massμ.

μ=

mcmπ

mc+mπ

=

(12× 1 ,840)(270)

[(12× 1 ,840)+(270)]

= 266. 7 me

For eachπ−,E=− 13. 6 × 266. 7 = 3 ,628 eV

For the 6 pions,E= 3 , 628 × 6 = 21 ,766 eV