75. A generalized formation reaction can be written X + x

→ Y, where X is the target nucleus, x is the

incident light particle, and Y is the excited compound nucleus (

20

Ne). We assume X is initially at rest.

Then, conservation of energy yields

m

X

c

2

+ m

x

c

2

+ K

x

= m

Y

c

2

+ K

Y

+ E

Y

where m

X

, m

x

, and m

Y

are masses, K

x

and K

Y

are kinetic energies, and E

Y

is the excitation energy

of Y. Conservation of momentum yields

p

x

= p

Y

.

Now, K

Y

= p

2
Y

/2m

Y

= p

2

x

/2m

Y

= (m

x

/m

Y

)K

x

, so

m

X

c

2

+ m

x

c

2

+ K

x

= m

Y

c

2

+ (m

x

/m

Y

)K

x

+ E

Y

and

K

x

=

m

Y

m

Y

− m

x

(m

Y

− m

X

− m

x

)c

2

+ E

Y



.

(a)Let x represent the alpha particle and X represent the

16

O nucleus. Then, (m

Y

− m

X

− m

x

)c

2

=

(19.99244 u

− 15.99491 u − 4.00260 u)(931.5 MeV/u)= −4.722 MeV and

K

α

=

19.99244 u

19.99244 u

− 4.00260 u

(

−4.722 MeV + 25.0 MeV) = 25.35 MeV .

(b)Let x represent the proton and X represent the

19

F nucleus.

Then, (m

Y

− m

X

− m

x

)c

2

=

(19.99244 u

− 18.99841 u − 1.00783 u)(931.5 MeV/u)= −12.85 MeV and

K

α

=

19.99244 u

19.99244 u

− 1.00783 u

(

−12.85 MeV + 25.0 MeV) = 12.80 MeV .

(c)Let x represent the photon and X represent the

20

Ne nucleus. Since the mass of the photon is

zero, we must rewrite the conservation of energy equation: if E

γ

is the energy of the photon, then

E

γ

+ m

X

c

2

= m

Y

c

2

+ K

Y

+ E

Y

. Since m

X

= m

Y

, this equation becomes E

γ

= K

Y

+ E

Y

. Since

the momentum and energy of a photon are related by p

γ

= E

γ

/c, the conservation of momentum

equation becomes E

γ

/c = p

Y

. The kinetic energy of the compound nucleus is K

Y

= p

2
Y

/2m

Y

=

E

2

γ

/2m

Y

c

2

. We substitute this result into the conservation of energy equation to obtain

E

γ

=

E

2

γ

2m

Y

c

2

+ E

Y

.

This quadratic equation has the solutions

E

γ

= m

Y

c

2

±



(m

Y

c

2

)

2

− 2m

Y

c

2

E

Y

.

If the problem is solved using the relativistic relationship between the energy and momentum of
the compound nucleus, only one solution would be obtained, the one corresponding to the negative
sign above. Since m

Y

c

2

= (19.99244 u)(931.5 MeV/u)= 1.862

× 10

4

MeV,

E

γ

=

(1.862

× 10

4

MeV)

−



(1.862

× 10

4

MeV)

2

− 2(1.862 × 10

4

MeV)(25.0 MeV)

=

25.0 MeV .

The kinetic energy of the compound nucleus is very small; essentially all of the photon energy goes
to excite the nucleus.