A photon has momentum of magnitude kg-m/s. What is the energy of this photon? Give your answer in joules and in electron volts.
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0. Math Review
Math Review
Problem 32
Textbook Question
In an experiment done in a laboratory on the earth, the wavelength of light emitted by a hydrogen atom in the to transition is nm. In the light emitted by the quasar 3C273 (see Problem ), this spectral line is redshifted to nm. Assume the redshift is described by Eq. () and use the Hubble law to calculate the distance in light-years of this quasar from the earth.

1
Step 1: Understand the redshift phenomenon. Redshift occurs when the wavelength of light emitted by an object is stretched due to the object moving away from the observer. The redshift (z) is defined as \( z = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} \), where \( \lambda_{observed} \) is the observed wavelength and \( \lambda_{emitted} \) is the emitted wavelength.
Step 2: Calculate the redshift \( z \) using the given wavelengths. Substitute \( \lambda_{observed} = 563.9 \, \text{nm} \) and \( \lambda_{emitted} = 486.1 \, \text{nm} \) into the formula \( z = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} \).
Step 3: Use the Hubble law to relate the redshift to the distance of the quasar. The Hubble law states \( v = H_0 \cdot d \), where \( v \) is the recession velocity, \( H_0 \) is the Hubble constant, and \( d \) is the distance. The recession velocity \( v \) can be approximated as \( v \approx c \cdot z \), where \( c \) is the speed of light.
Step 4: Substitute \( v = c \cdot z \) into the Hubble law to find the distance \( d \). Rearrange the equation to \( d = \frac{v}{H_0} \), and substitute \( v \approx c \cdot z \) to get \( d = \frac{c \cdot z}{H_0} \). Use the known values for \( c \) (speed of light) and \( H_0 \) (Hubble constant, typically \( 70 \, \text{km/s/Mpc} \) or another given value).
Step 5: Convert the distance \( d \) from megaparsecs (Mpc) to light-years. Use the conversion factor \( 1 \, \text{Mpc} = 3.26 \times 10^6 \, \text{light-years} \). Multiply the distance in Mpc by this factor to express the final distance in light-years.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Wavelength and Energy Levels
In atomic physics, the wavelength of light emitted during electron transitions between energy levels is crucial. For hydrogen, when an electron moves from a higher energy level (n=4) to a lower one (n=2), it emits a photon with a specific wavelength, which can be calculated using the Rydberg formula. This relationship between energy levels and emitted wavelengths is foundational for understanding atomic spectra.
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Redshift
Redshift occurs when light from an object is stretched to longer wavelengths as it moves away from an observer, often due to the expansion of the universe. It is quantified by the formula z = (λ_observed - λ_emitted) / λ_emitted, where λ represents the wavelengths. In cosmology, redshift is a key indicator of the velocity and distance of celestial objects, such as quasars.
Hubble's Law
Hubble's Law states that the recessional velocity of a galaxy (v) is directly proportional to its distance (d) from Earth, expressed as v = H₀ * d, where H₀ is the Hubble constant. This relationship allows astronomers to estimate distances to faraway galaxies and quasars based on their redshift, providing insights into the structure and expansion of the universe.
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