1000 Solved Problems in Modern Physics

448 8 Nuclear Physics – II

The mass of this ion is then

9. 98 × 10 −^27 kg

1. 66 × 10 −^27 kg/amu

= 6 .012 amu

Therefore the mass number is 6.

For the second ion, the only change is the radius of the orbit which is

0.0954 m. The mass of the second ion is

m 2 =m 1 ×

(r 2

r

) 2

= 6. 012 ×

(

0. 0954

0. 0883

) 2

= 7 .0176 amu

Therefore the mass number is 7.

8.2 The radius of curvature of an ion in the magnetic inductionB, perpendicular to
the orbit, with kinetic energyqVwill be

r=

(

2 MV

qB^2

) 1 / 2

(1)

After a deviation of 180◦the two ions will be separated byd, the difference

between the diameters of the circular path.

(a)d=2(r 11 −r 10 )=

(

8 V

qB^2

) 1 / (^2) (√
M 1 −

√

M 2

)

(2)

SubstituteV= 5 ,000 V,q= 1. 6 × 10 −^19 C,B= 0 .15 T, M 1 = 11 ×

1. 66 × 10 −^27 Kg andM 2 = 10 × 1. 66 × 10 −^27 Kg to find the separation

d= 0 .021 m or 2.1 cm.

(b) From (1), we get

ΔM

M

= 2

Δr

r

The mass resolution is

δ=

M

ΔM

=

r

2 Δr

=

r

d

=

22 cm

2 .1cm

= 10. 5

whereris the mean radius of the ions,r 11 andr 10 , determined from (1) as

22.52 cm and 21.47 cm.

8.3 p=

√

2 Tm=qBr

HereT, qandrare fixed.

∴

B 2

B 1

=

(

m 2

m 1

)

=

37

35

Increase in induction

ΔB=B 2 −B 1 =B 1

(

37

35

− 1

)

=

2 B 1

35

=

2 × 0. 1

35

= 0. 57 × 10 −^3 Tor5.7G

8.4^27 Si→^27 Al+β++ν+Tmax

MSi−MAl= 2 me+Tmax= 2 × 0. 511 + 3. 48 = 4 .5MeV

The transition is between two mirror nuclei of chargeZ+1 andZ.The

difference in Coulomb energy is