Using de Broglie’s approach, ?nd the allowed energy levels and the electron radii
for a hydrogen–like atom, i.e. one in which a single electron orbits a nucleus of
charge +Ze, where Z is the atomic number. Z = 1 then gives the hydrogen atom
itself.

1)Find the lowest (n = 1) energy level and orbital radius for the case Z = 92.
See if you can ?nd an estimate for the radius of the U nucleus, and compare
with this orbital radius.

2)Angular momentum in the stellar systems: as discussed in the context of the
virial theorem, a protostellar system must lose energy in order to coalesce,
condense, and heat up. Furthermore, it must also shed angular momentum.
A gas cloud of stellar mass will initially possess enormous angular momen-tum even if it is rotating very slowly, from random motion of its constituent
material. Some protostellar systems solve, or partially solve, this problem by evolving into close binary stars. Others probably do so by “growing” a plan-etary system. We are accustomed to think of the planets in our own solar
system as inconsequential in relation to the Sun — negligible in terms of mass
and in terms of power production. However, the orbital motion of the planets
accounts for well over 90% of the total angular momentum of the solar system,
as the following calculations will show.

i: Assume a ball of H gas of uniform density, 500 atoms/cm3, one light–
year in radius, rotating uniformly about its centre so that its equator is
moving at the very moderate speed of 1 m
s- 1. Compute its mass and
its angular momentum.

It is an oversimpli?cation to suppose that a cloud of interstellar gas will ever ro-tate uniformly, but this calculation gives the order of magnitude of the amount
of angular momentum which such a cloud can possess.

ii: Assuming that the Sun is a Standard Model star rotating uniformly, com-pute its angular momentum. Take its period to be 30 days.
In fact the Sun rotates somewhat more rapidly at its equator (period 24.6 days)
than near its poles (period 34 days).

iii: Calculate the total orbital angular momentum of Jupiter, Saturn, Uranus
and Neptune around the Sun.

iv: Compare the three angular momenta from i, ii, and iii.

v: One of the largest angular velocities known among stars is exhibited
by Altair ( Aquilæ, the 12th brightest star in the sky), with a period
of approximately 6.5 hours. How would that period have been deter-mined? Assuming that Altair can be described by the Standard Model,
with M = 1.6M and R = 1.77R , calculate its rotational angular mo-mentum.

Altair is of spectral type A7 V; stars earlier than about F5 rotate, usually, much
faster than the later stars. This has been interpreted to mean that early Main
Sequence stars, for whatever reason, are less prone to “growing” planetary sys-tems than late MS stars.
The moment of inertia of a uniform ball of radius R and mass M about its centre is
( 2=5) M R2
 while the moment of inertial of a Standard Model star is 0.0754 M R2
 . Data
on planetary masses and orbital parameters can be found in Carroll and Ostlie, or in
Wikipedia.

To tackle the question regarding the allowed energy levels and electron radii for a hydrogen-like atom using de Broglie's approach, we need to start by applying the principles of quantum mechanics and classical physics. Let's break this down step by step.

Energy Levels and Electron Radii for Hydrogen-like Atoms

For a hydrogen-like atom, the energy levels can be derived from the Bohr model, which is based on de Broglie's hypothesis that particles exhibit wave-like properties. The energy levels for an electron in a hydrogen-like atom are given by the formula:

E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2}

Here, Z is the atomic number, n is the principal quantum number (1, 2, 3,...), and 13.6 eV is the ionization energy of hydrogen (Z=1).

Calculating for Uranium (Z = 92)

For the lowest energy level (n = 1) in a uranium atom:

E_1 = -\frac{92^2 \cdot 13.6 \text{ eV}}{1^2} = -\frac{8464 \cdot 13.6 \text{ eV}}{1} = -115,584 \text{ eV}

This indicates that the energy required to remove the electron from the ground state of a uranium atom is approximately 115.6 keV.

Finding the Orbital Radius

The radius of the electron's orbit can be calculated using the formula:

r_n = \frac{n^2 \cdot 4 \pi \epsilon_0 \hbar^2}{Z \cdot e^2}

Where:

• ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 C²/(N·m²))
• ħ is the reduced Planck's constant (approximately 1.055 x 10^-34 J·s)
• e is the elementary charge (approximately 1.602 x 10^-19 C)

Substituting for n = 1 and Z = 92:

r_1 = \frac{1^2 \cdot 4 \pi (8.85 \times 10^{-12}) (1.055 \times 10^{-34})^2}{92 \cdot (1.602 \times 10^{-19})^2}

Calculating this gives:

r_1 ≈ 2.5 \times 10^{-14} \text{ m} = 25 \text{ pm}

Estimating the Radius of the Uranium Nucleus

The radius of a nucleus can be estimated using the formula:

R = R_0 \cdot A^{1/3}

Where R₀ is approximately 1.2 fm (femtometers) and A is the mass number (for uranium, A ≈ 238).

Calculating gives:

R ≈ 1.2 \cdot 238^{1/3} ≈ 7.1 \text{ fm}

Comparing the two radii, the orbital radius of the electron (25 pm) is significantly larger than the nuclear radius (7.1 fm), which aligns with our understanding that electrons occupy much larger regions compared to the size of the nucleus.

Angular Momentum in Stellar Systems

Now, let's shift our focus to angular momentum in stellar systems, particularly in protostellar clouds and their evolution.

Angular Momentum of a Gas Cloud

To compute the mass of a gas cloud with a uniform density of 500 atoms/cm³ and a radius of one light-year:

First, convert the radius to centimeters:

1 \text{ light-year} = 9.461 \times 10^{17} \text{ cm}

The volume of the cloud is:

V = \frac{4}{3} \pi r^3

Calculating the volume:

V ≈ \frac{4}{3} \pi (9.461 \times 10^{17})^3 \approx 3.54 \times 10^{53} \text{ cm}^3

The mass of the gas cloud can be estimated by multiplying the volume by the density:

M = V \cdot \text{density} = 3.54 \times 10^{53} \text{ cm}^3 \cdot 500 \text{ atoms/cm}^3 \cdot 1.67 \times 10^{-24} \text{ g/atom} ≈ 2.95 \times 10^{30} \text{ g}

Next, we calculate the angular momentum (L) of the cloud, assuming it rotates uniformly:

L = I \cdot \omega

Where I is the moment of inertia and ω is the angular velocity. For a uniform sphere:

I = \frac{2}{5} M R^2

Assuming a very slow rotation speed (1 m/s), we can find ω:

ω = \frac{v}{R} = \frac{1 \text{ m/s}}{9.461 \times 10^{17} \text{ cm} \cdot 10^{-2}} ≈ 1.06 \times 10^{-19} \text{ rad/s}

Now substituting into the angular momentum formula gives:

L ≈ \frac{2}{5} (2.95 \times 10^{30}) (9.461 \times 10^{17} \cdot 10^{-2})^2 (1.06 \times 10^{-19})

This calculation