"Atoms and Nuclei" — Practice

Question 1

An alpha particle of $5.5$ MeV is fired head-on at a silver nucleus ($Z=47$). Find the distance of closest approach, and compare it with the nuclear radius given $A=108$.

Accepted Answer

$r_0=\dfrac{2	imes47	imes1.44}{5.5}=\dfrac{135.4}{5.5}=24.6$ fm

$R=1.2	imes108^{1/3}=1.2	imes4.76=5.7$ fm

The alpha stops more than four nuclear radii away, so it never enters the nucleus and the analysis is purely electrostatic — exactly the condition under which Rutherford's formula is valid.

Question 2

Find the radius, total energy and orbital speed of the $n=3$ state of the $	ext{He}^{+}$ ion.

Accepted Answer

$r=\dfrac{0.529	imes9}{2}=2.38$ Å

$E=-\dfrac{13.6	imes4}{9}=-6.04$ eV

$v=\dfrac{c}{137}	imes\dfrac23=1.46	imes10^{6}$ m s$^{-1}$

The kinetic energy is $+6.04$ eV, since the total energy of a bound Coulomb orbit is exactly minus the kinetic energy.

Question 3

Show that the ratio of the de Broglie wavelength of the electron to the circumference of its Bohr orbit is $1/n$, and interpret the result.

Accepted Answer

Bohr: $mvr_n=\dfrac{nh}{2\pi}$, so $mv=\dfrac{nh}{2\pi r_n}$.

$\lambda=\dfrac{h}{mv}=\dfrac{2\pi r_n}{n}$

$\dfrac{\lambda}{2\pi r_n}=\dfrac1n$

Exactly $n$ wavelengths fit around the $n$th orbit. The quantisation is therefore the condition for a **standing wave**, not an arbitrary postulate.

Question 4

Hydrogen atoms in the ground state are bombarded by electrons of $12.5$ eV. Find which levels can be excited and how many spectral lines the gas will emit.

Accepted Answer

Excitation energies from $n=1$: to $n=2$ needs $10.2$ eV, to $n=3$ needs $12.09$ eV, to $n=4$ needs $12.75$ eV.

With $12.5$ eV available, levels up to $n=3$ can be reached but not $n=4$.

Lines $=\dfrac{3	imes2}{2}=3$: the transitions $3	o2$, $3	o1$ and $2	o1$.

Question 5

Find the shortest and longest wavelengths of the Lyman series of hydrogen.

Accepted Answer

Longest is $n=2	o1$: $\Delta E=13.6-3.40=10.2$ eV, so $\lambda=\dfrac{1240}{10.2}=121.6$ nm.

Shortest is $n=\infty	o1$: $\Delta E=13.6$ eV, so $\lambda=\dfrac{1240}{13.6}=91.2$ nm.

Both lie in the ultraviolet, which is why the Lyman series is invisible to the eye and was found only with a spectrograph.

Question 6

Estimate the wavelength difference between the red Balmer lines of hydrogen and deuterium, given that the reduced-mass correction changes the Rydberg constant by one part in $3700$.

Accepted Answer

The hydrogen line is at $656.3$ nm.

$\dfrac{\Delta\lambda}{\lambda}=\dfrac{1}{3700}$

$\Delta\lambda=\dfrac{656.3}{3700}=0.177$ nm

The deuterium line lies about $0.18$ nm to the **short-wavelength** side, since its heavier nucleus gives a larger reduced mass and hence a larger Rydberg constant.

Question 7

Find the binding energy per nucleon of $^{16}$O, given atomic masses $m_H=1.00783$ u, $m_n=1.00867$ u and $M=15.99491$ u.

Accepted Answer

$\Delta m=8\left(1.00783ight)+8\left(1.00867ight)-15.99491$

$=8.06264+8.06936-15.99491=0.13709$ u

$BE=0.13709	imes931.5=127.7$ MeV

$\dfrac{BE}{A}=\dfrac{127.7}{16}=7.98$ MeV, just below the peak value near iron.

Question 8

Compare the radii and densities of $^{27}$Al and $^{216}$Rn nuclei.

Accepted Answer

$R_{Al}=1.2	imes27^{1/3}=1.2	imes3=3.6$ fm

$R_{Rn}=1.2	imes216^{1/3}=1.2	imes6=7.2$ fm

The radon nucleus is exactly **twice** as large, since $216/27=8$.

Its mass is eight times greater and its volume is eight times greater, so the two densities are **identical** — which is the general result for all nuclei.

Question 9

A $^{235}$U nucleus with binding energy per nucleon $7.6$ MeV splits into fragments of mass numbers $140$ and $95$, both with binding energy per nucleon $8.5$ MeV. Estimate the energy released.

Accepted Answer

Before: $235	imes7.6=1786$ MeV

After: $\left(140+95ight)	imes8.5=235	imes8.5=1998$ MeV

$Q=1998-1786=212$ MeV

The fragments are more tightly bound, and the difference appears mostly as their kinetic energy. About $200$ MeV per fission is the standard figure.

Question 10

A radioactive sample has a half-life of $20$ minutes. Find the fraction remaining after $50$ minutes, and the time for the activity to fall to $10\%$ of its initial value.

Accepted Answer

$\lambda=\dfrac{0.693}{20}=0.03466$ per minute

After $50$ min: $\dfrac{N}{N_0}=e^{-0.03466	imes50}=e^{-1.733}=0.177$, about $17.7\%$.

For $10\%$: $\lambda t=\ln10=2.303$

$t=\dfrac{2.303}{0.03466}=66.4$ minutes, a little over three half-lives.

Question 11

Thorium-232 ($T_{1/2}=1.4	imes10^{10}$ y) decays through a chain to radium-228 ($T_{1/2}=5.75$ y). Find the ratio of radium to thorium atoms in an old mineral, and the ratio of their activities.

Accepted Answer

$\dfrac{N_{Ra}}{N_{Th}}=\dfrac{T_{Ra}}{T_{Th}}=\dfrac{5.75}{1.4	imes10^{10}}=4.1	imes10^{-10}$

In secular equilibrium the two **activities are equal**, so their ratio is exactly $1$.

That is the defining feature of the equilibrium: vastly fewer daughter atoms, each decaying vastly faster, giving the same number of decays per second.

Question 12

A $^{232}$Th nucleus decays to $^{208}$Pb. Find the numbers of alpha and beta-minus decays, given $Z=90$ for thorium and $Z=82$ for lead.

Accepted Answer

Mass number falls by $232-208=24$, so alphas $=\dfrac{24}{4}=6$.

Six alphas alone would take $Z$ from $90$ to $90-12=78$.

The actual final value is $82$, so beta-minus decays must raise it by $4$.

Beta-minus decays $=4$.

Practice — "Atoms and Nuclei"

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