Question 1

A planetary nebula is created:

Accepted Answer

A planetary nebula is created over a period of 10,000 years or more due to thermal pulses in a low-mass star. It is the result of the outer layers of the star being expelled into space as the core contracts and heats up. This process occurs in the final stages of the star's life, after it has exhausted the available nuclear fuel in its core. Eventually, the remaining core of the star will become a white dwarf.

Question 2

The major source of light in the expanding shell of gas in a planetary nebula is:

Accepted Answer

The major source of light in a planetary nebula is fluorescence of the atoms, which is caused by UV radiation from the hot central white dwarf star. As the radiation passes through the expanding shell of gas, it ionizes the gas atoms, causing them to emit light in specific wavelengths. This results in the bright, colorful appearance of planetary nebulae that we observe.

Question 3

The typical diameter of a planetary nebula is:

Accepted Answer

Planetary nebulae typically have diameters of about 1 light-year. They are formed from the outer layers of a star that has shed its material towards the end of its life cycle, expanding over time as they move away from the central star.

Question 4

The core of a planetary nebula has an initial surface temperature of 100,000 K. If it radiates like a blackbody, in what electromagnetic region will the peak of its radiation spectrum lie?

Accepted Answer

The peak of radiation emitted by a blackbody is given by Wien's Law, which states that λpeak = 2.898 × 10⁶ / T, where λpeak is the wavelength of the peak emission and T is the temperature in Kelvin. Substituting T = 100,000 K gives λpeak ≈ 29 nm, which corresponds to the ultraviolet region. Therefore, the best choice is C, ultraviolet.

Question 5

The fraction of the mass of a red giant that is ejected as a shell in a planetary nebula is:

Accepted Answer

Red giants can eject up to 40% of their mass as they transition into planetary nebulae. This process is part of the star's evolution towards becoming a white dwarf, where the outer layers are expelled, leaving behind the core.

Question 6

What will be the mass of the Sun at the end of its asymptotic giant branch (AGB) phase, as a result of mass loss to space by its stellar wind?

Accepted Answer

The Sun is expected to lose a significant portion of its mass due to stellar wind during its AGB phase, leaving behind a core that is about 0.6 times the mass of the Sun (0.6 M☉). This mass loss is a critical part of the evolution of low to intermediate-mass stars like the Sun.

Question 7

The shell of a planetary nebula is measured by the Doppler shift of emission lines to be expanding outward at a speed of 10 km/s and its diameter is measured to be 1 ly. How long has the shell been expanding?

Accepted Answer

The shell has been expanding for 50,000 years. This is calculated by dividing the diameter of the nebula (1 light-year) by the expansion speed (10 km/s). Since 1 light-year is the distance light travels in one year, and the speed of light is approximately 300,000 km/s, we convert the expansion speed to the same units and find the time it takes to cover that distance.

Question 8

Out of all matter ejected into the interstellar medium each year by stars, planetary nebula shell ejections account for:

Accepted Answer

The answer of Out of all matter ejected into the...

Question 9

What fraction of the material returned to the interstellar medium of a galaxy by stars is contributed by planetary nebulae as they eject their shells?

Accepted Answer

Planetary nebulae contribute about 15% of the material returned to the interstellar medium of a galaxy as they eject their shells. This process occurs when stars of intermediate mass shed their outer layers at the end of their life cycles.

Question 10

Which of the following important components does a planetary nebula contribute to the interstellar medium?

Accepted Answer

Planetary nebulae release nuclei of elements like carbon, nitrogen, and oxygen into the interstellar medium, enriching it with materials that are essential for the formation of new stars and planets.

Question 11

What causes thermal pulses in low-mass stars?

Accepted Answer

Thermal pulses in low-mass stars are caused by a helium shell flash, which ignites the helium shell and triggers an expansion of the outer layers.

Question 12

The energy to produce a thermal pulse comes from:

Accepted Answer

A thermal pulse occurs during the asymptotic giant branch (AGB) phase of a star's evolution, when the helium shell around the core ignites explosively. This release of energy creates a shell of hot gas that expands outward, causing the star's outer envelope to expand and cool. The energy for this process comes from the gravitational energy stored in the core as it contracts and heats up, which triggers the helium shell flash. Therefore, the correct choice is D - the helium shell flash.

Question 13

The object seen at the center of a planetary nebula is:

Accepted Answer

The central object of a planetary nebula is the remnant core of a red giant star, which has collapsed into a white dwarf star.

Question 14

According to our theories of stellar evolution, the Sun will probably end its active life as a:

Accepted Answer

According to our theories of stellar evolution, the Sun, which is a low-mass star, will eventually exhaust its nuclear fuel and expand into a red giant. Once it exhausts its outer envelope, it will shrink down to become a white dwarf. It does not have enough mass to form a black hole or a pulsar.

Question 15

In which order will a single star of about 1 M ? progress through its various stages of evolution?

Accepted Answer

The correct sequence for a star of about 1 solar mass (1 M☉) starts with its formation phase as a T Tauri star, then it enters the main sequence where it spends most of its life fusing hydrogen into helium. After exhausting its hydrogen fuel, the star expands into a red giant (though not explicitly mentioned in option A, this phase occurs before the planetary nebula stage), then sheds its outer layers to form a planetary nebula, and finally, the core that remains cools down to become a white dwarf.

Question 16

One particular characteristic of a white dwarf star is its:

Accepted Answer

White dwarf stars have a very high surface temperature, usually around 15,000K to 100,000K. This is because they are the leftover cores of stars that have exhausted their nuclear fuel and collapsed.

Question 17

A white dwarf star is supported from collapse under gravity by:

Accepted Answer

A white dwarf star is formed after a low to medium mass star has exhausted all its nuclear fuel and the outer layers drift off into space, leaving behind a core primarily composed of carbon and oxygen. These atoms are packed closely together and are mostly ionized and therefore, no longer bound by electron bonds, and they exert pressure at the same time trying to collapse as . Under the Pauli Exclusion Principle, which states that no two fermions (in this case, electrons) can occupy the same quantum state simultaneously, the electrons resist the pressure from gravity and provide electron degeneracy pressure to support the star from further collapse.

Question 18

When the small but dense white dwarf is formed from a star like the Sun, the core temperature is not high enough to permit the carbon and oxygen to participate in further energy-producing nuclear reactions. But why does this dense core not produce energy by the Kelvin-Helmholtz contraction?

Accepted Answer

The core of the white dwarf is supported by electron degeneracy pressure, which resists further compression and contraction. This pressure arises from the fact that electrons, being fermions, cannot occupy the same quantum state and must fill higher energy states as the core compresses. As the number of electrons per unit volume increases, their average kinetic energy also increases, generating a pressure that holds up the core against further contraction. Therefore, the Kelvin-Helmholtz contraction cannot occur in a white dwarf.

Question 19

As a red giant evolves, its temperature decreases but its luminosity increases. As a white dwarf ages, its temperature decreases, and its luminosity also decreases. Why this difference?

Accepted Answer

The expansion of a red giant as it cools is due to the decrease in temperature causing the outer layers of the star to become less tightly bound to the core. On the other hand, a white dwarf is supported by degeneracy pressure, which does not depend on temperature, so it maintains the same size as it cools. This means that a white dwarf's luminosity decreases as its temperature decreases, unlike a red giant, where the luminosity increases as the star expands.

Question 20

Because it has ceased nuclear burning in its interior and therefore no longer generates energy, why is it that a white dwarf does not shrink rapidly under the force of gravity as it cools?

Accepted Answer

The electrons within a white dwarf are in a degenerate state, which means they follow the rules of quantum mechanics rather than classical physics. According to the Pauli exclusion principle, no two electrons can occupy the same energy state. This means that as a white dwarf cools and its electrons lose energy, they cannot collapse into the same energy state and must occupy higher and higher energy states. This creates a strong electron degeneracy pressure that opposes the force of gravity, preventing the white dwarf from shrinking rapidly.

Quiz 20: Stellar Evolution: The Death of Stars