Problems

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Calculate the total binding energy of ²⁰₁₀ Ne.

Solution:
By Table 29.3 in the text, we see that ²⁰₁₀ Ne has a mass of 19.992439 u. Since Neon has 10 protons and 10 neutrons, the mass deficiency is

10 x 1.007825 + 10 x 1.008665 - 19.992439 = 0.172461 u .

This corresponds to an energy of

E = mc ² = 0.172461 u x 931.5 $${\frac{{\:\rm MeV}}{{\:\rm u}\ {\:\rm c}^2}}$$ c ² = 160.6 MeV .

Thus, the binding energy is 160.6 MeV, or about 8.0 MeV per nucleon.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Suppose you begin with 1.0 x 10^{- 2} g of a pure radioactive substance and 4 h later determine that 0.25 x 10^{- 2} g remain. What is the half-life of the substance?

Solution:
For this we will use the relation

N = N₀e ^{- $\lambda$t} $$\Rightarrow$$ $${\frac{N}{N_0}}$$ = $$\left(\frac{1}{2}\right)$$^t/T_1/2,

where T_1/2 = ln 2/$\lambda$ has been used. We then have

$${\frac{0.25\times 10^{-2}}{1.0\times 10^{-2}}}$$ = 0.25 = $$\left(\frac{1}{2}\right)$$^t/T_1/2 $$\Rightarrow$$ ln 0.25 = $${\frac{4\ {\:\rm h}}{T_{1/2}}}$$ln $$\left(\frac{1}{2}\right)=$$ - $${\frac{4\ {\:\rm h}}{T_{1/2}}}$$ln 2,

from which we find T_1/2 = 2.0 h.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

The ¹⁴C content decreases after the death of a living system with a half-life of 5739 years. If the ¹⁴C content of an old piece of wood is found to be 12.5% of that of an equivalent present-day sample, how old is the piece of wood?

Solution:
This will also use the relation

N = N₀e ^{- $\lambda$t} $$\Rightarrow$$ $${\frac{N}{N_0}}$$ = $$\left(\frac{1}{2}\right)$$^t/T_1/2.

With the data given, we find

0.125 = $$\left(\frac{1}{2}\right)$$^{t/5739 yr} $$\Rightarrow$$ ln 0.125 = $${\frac{t}{5739\ {\:\rm yr}}}$$ln $$\left(\frac{1}{2}\right)=$$ - $${\frac{t}{5739\ {\:\rm yr}}}$$ ln 2,

from which we deduce that t = 17,217 yr.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Find the energy released in the alpha-decay

²³⁸₉₂ U $$\to$$²³⁴₉₀ Th + ⁴₂ He

Solution:
For this we shall need the masses

$$\left({}^{238}_{92}{\:\rm U}\right)=$$

$$\left({}^{234}_{90}{\:\rm Th}\right)=$$

$$\left({}^{4}_{22}{\:\rm He}\right)=$$

from which we find the mass difference to be

238.050786 - (234.043583 + 4.002603) = 0.0046 u .

This corresponds to an energy of

E = mc ² = 0.0046 u x 931.5 $${\frac{{\:\rm MeV}}{{\:\rm u}\ {\:\rm c}^2}}$$ c ² = 4.29 MeV .

Thus, the reaction releases an energy of 4.29 MeV.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Suppose that the sun consists entirely of hydrogen and that the dominant energy-releasing reaction is

4$$\left({}^{1}_{1}{\:\rm H}\right)\to$$⁴₂ He + 2$$\left({}^{0}_{1}{\:\rm e}\right)+$$2$$\nu$$ + $$\gamma$$.

If the total power output of the sun is assumed to remain constant at 3.9 x 10²⁶ W, how long will it take for all of the hydrogen to be burned up? Take the mass of the sun as 1.99 x 10³⁰ kg.

Solution:
We first find the total number of hydrogen atoms in the sun by calculating

$${\frac{1.99\times 10^{30} \ {\:\rm kg}}{1.67\times 10^{-27}\ {\:\rm kg}/\ {\:\rm atom}}}$$ = 1.192 x 10⁵⁷ atoms .

Now, the reaction quoted has a mass difference of

4 x 1.007825 - (4.002603 + 2 x 0.000549) = 0.027599 u ,

which corresponds to an energy of

E = mc ² = 0.027599 u x 931.5 $${\frac{{\:\rm MeV}}{{\:\rm u}\ {\:\rm c}^2}}$$ c ² = 25.71 MeV .

Since each individual reaction consumes 4 Hydrogen atoms, the total energy available in the sun is

1.192 x 10⁵⁷ atoms x 25.71 $${\frac{{\:\rm MeV}}{4\ {\:\rm atoms}}}$$ x 10⁶$${\frac{{\:\rm eV}}{{\:\rm MeV}}}$$ x $${\frac{1.6\times 10^{-19}\ {\:\rm J}}{{\:\rm eV}}}$$ = 1.225 x 10⁴⁵ J .

The lifetime of the sun can then be estimated as

$${\frac{1.225\times 10^{45}\ {\:\rm J}}{3.9\times 10^{26}\ {\:\rm J}/\ {\:\rm s}}}$$ x $${\frac{1 \ {\:\rm h}}{3600\ {\:\rm s}}}$$ x $${\frac{1 \ {\:\rm d}}{24\ {\:\rm h}}}$$ x $${\frac{1 \ {\:\rm yr}}{365\ {\:\rm d}}}$$ = 9.96 x 10¹⁰ yr .

Thus, this gives an estimated lifetime of about 99.6 billion years.