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(a) What is the wavelength of an X-ray photon of energy \(10.0 \mathrm{keV} ?(b)\) What is the wavelength of a gamma-ray photon of energy \(1.00 \mathrm{MeV} ?\) (c) What is the range of\begin{aligned} &\text { energies of photons of visible light with wavelengths }\\\ &350-700 \mathrm{nm} ? \end{aligned}

Short Answer

Expert verified
(a) 1.24 x 10^{-10} m, (b) 1.24 x 10^{-12} m, (c) 1.775 eV - 3.55 eV

Step by step solution

01

Understanding the relationship between energy and wavelength

The energy of a photon is given by the equation: \[ E = \frac{hc}{\text{λ}} \]where \( E \) is the energy, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{J} \text{s}) \), \( c \) is the speed of light \( (3.00 \times 10^{8} \text{m/s}) \), and \( \text{λ} \) is the wavelength.
02

Convert energy to joules

For X-ray photon: Convert 10.0 keV to joules: \[ 10 \text{keV} = 10 \times 10^3 \times 1.602 \times 10^{-19} \text{J} = 1.602 \times 10^{-15} \text{J} \] For gamma-ray photon: Convert 1.00 MeV to joules: \[ 1 \text{MeV} = 1 \times 10^6 \times 1.602 \times 10^{-19} \text{J} = 1.602 \times 10^{-13} \text{J} \]
03

Calculate wavelengths

For the X-ray photon: Use \( \text{λ} = \frac{hc}{E} \)\[ \text{λ}_{\text{X-ray}} = \frac{6.626 \times 10^{-34} \text{J} \text{s} \times 3.00 \times 10^{8} \text{m/s}}{1.602 \times 10^{-15} \text{J}} = 1.24 \times 10^{-10} \text{m} \]For the gamma-ray photon: \[ \text{λ}_{\text{gamma}} = \frac{6.626 \times 10^{-34} \text{J} \text{s} \times 3.00 \times 10^{8} \text{m/s}}{1.602 \times 10^{-13} \text{J}} = 1.24 \times 10^{-12} \text{m} \]
04

Determine the energy range for visible light

Convert the given wavelengths to meters: \[ 350 \text{nm} = 350 \times 10^{-9} \text{m}, \text{ and } 700 \text{nm} = 700 \times 10^{-9} \text{m} \]For the shortest wavelength (highest energy): \[ E_{\text{max}} = \frac{hc}{\text{λ}_{\text{min}}} = \frac{6.626 \times 10^{-34} \text{J} \text{s} \times 3.00 \times 10^{8} \text{m/s}}{350 \times 10^{-9} \text{m}} = 5.68 \times 10^{-19} \text{J} = 3.55 \text{eV} \]For the longest wavelength (lowest energy): \[ E_{\text{min}} = \frac{hc}{\text{λ}_{\text{max}}} = \frac{6.626 \times 10^{-34} \text{J} \text{s} \times 3.00 \times 10^{8} \text{m/s}}{700 \times 10^{-9} \text{m}} = 2.84 \times 10^{-19} \text{J} = 1.775 \text{eV}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

X-ray photon energy
X-ray photons pack a lot of energy. They have energies typically in the range of hundreds of electron volts (eV) to several kiloelectron volts (keV). One key feature of X-rays is their ability to penetrate materials. This makes them useful for medical imaging and security scanning. To understand X-ray photon energy, it's essential to grasp the relationship between energy and wavelength.
The formula for photon energy is: \[ E = \frac{hc}{\lambda} \]where \( E \) is energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. For instance, an X-ray photon with an energy of 10.0 keV (which needs to be converted into joules for calculations) can be used to find its wavelength. The relationship between energy and wavelength is inversely proportional, meaning as the energy increases, the wavelength decreases.
gamma-ray photon energy
Gamma rays are a form of electromagnetic radiation that have even more energy than X-rays. Gamma photons have energies exceeding mega electron volts (MeV). These high-energy photons can penetrate most materials, making them useful in cancer treatment and sterilizing medical equipment. To find the wavelength of a gamma-ray photon with 1.00 MeV energy, the same energy-wavelength relationship formula is applied: \[ E = \frac{hc}{\lambda} \] By converting the MeV value to joules, we can derive the wavelength. Gamma-ray photons have very short wavelengths due to their high energy levels, often around the picometer scale.
visible light wavelength
Visible light occupies a tiny portion of the electromagnetic spectrum but is crucial as it allows us to see. The wavelengths of visible light range from approximately 350 nm (violet) to 700 nm (red). These wavelengths correspond to energies that fall within a range that is much lower than those of X-rays and gamma rays. The energies of visible light photons can be calculated using the same formula:\[ E = \frac{hc}{\lambda} \] For instance, photons with a wavelength of 350 nm have more energy than those with a wavelength of 700 nm. This spectrum is essential in various applications, including photography, art, and everyday human vision.
Planck's constant
Planck's constant \((h)\) is a fundamental constant in physics, representing the quantized nature of energy. Its value is \(6.626 \times 10^{-34} \text{J} \text{s}\). This constant is crucial for understanding the energy of photons in relation to their frequency and wavelength.
In the equation \[ E = \frac{hc}{\lambda} \] \( h \) acts as the proportionality constant. It helps relate the energy (\(E\)) of a photon to its frequency (\( u \)) using another form of the equation: \[ E = hu \] With Planck's constant, it becomes possible to delve into quantum mechanics and the behavior of particles at very small scales.
energy-wavelength relationship
The energy-wavelength relationship is pivotal in physics. It states that the energy \(E\) of a photon is inversely proportional to its wavelength \(\lambda\), which can be expressed as: \[ E = \frac{hc}{\lambda} \] This means that as the wavelength decreases, the energy increases, and vice versa. This relationship is foundational for understanding various phenomena in the electromagnetic spectrum, from radio waves to gamma rays. It's also why different types of electromagnetic waves have varying effects. For example:
  • X-rays can penetrate skin and tissue, allowing doctors to see inside the body.
  • Gamma rays have enough energy to kill cancer cells.
  • Visible light enables us to see the colorful world around us.
This broad range of applications showcases the importance of the energy-wavelength relationship in both scientific theory and practical uses.

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Most popular questions from this chapter

A photon of energy \(E\) interacts with an electron at rest and undergoes pair production, producing a positive electron (positron) and an electron (in addition to the original electron): $$ \text { photon }+\mathrm{e}^{-} \rightarrow \mathrm{e}^{+}+\mathrm{e}^{-}+\mathrm{e}^{-} $$ The two electrons and the positron move off with identical momenta in the direction of the initial photon. Find the kinetic energy of the three final particles and find theenergy \(E\) of the photon. (Hint: Conserve momentum and total relativistic energy.)

A certain gamma-ray detector measures photon energies through the Compton interaction: the photon Compton scatters within the detector material, which then absorbs the kinetic energy of the scattered electron. The absorbed energy of the scattered electron is the response of the Suppose photons of energy \(E\) are incident on this detector. (a) Find an expression for the maximum energy response \(E_{\max }\) of this detector and show that \(E_{\max }\) is less than the original energy of the photon. (b) Evaluate \(E_{\max }\) for an incident photon energy of \(1.5 \mathrm{MeV}\). (c) Occasionally the detector may report events with energy greater than \(E_{\max }\) but less than \(E .\) What other processes might be responsible for such events? \((d)\) What processes might contribute to the detector reporting the full energy \(E\) of the photon?

(a) Assuming the Sun to radiate like an ideal thermal source at a temperature of \(6000 \mathrm{K}\), what is the intensity of the solar radiation emitted in the range \(530.0 \mathrm{nm}\) to \(532.0 \mathrm{nm} ?\) (b) What fraction of the total solar radiation does this represent?

When sodium metal is illuminated with light of wavelength \(4.20 \times 10^{2} \mathrm{nm},\) the stopping potential is found to be \(0.65 \mathrm{V} ;\) when the wavelength is changed to \(3.10 \times 10^{2} \mathrm{nm},\) the stopping potential is \(1.69 \mathrm{V}\). Using only these data and the values of the speed of light and the electronic charge, find the work function of sodium and a value of Planck's

The cutoff wavelength for the photoelectric effect in a certain metal is \(254 \mathrm{nm}\). (a) What is the work function for that metal? (b) Will the photoelectric effect be observed for \(\lambda>254 \mathrm{nm}\) or for \(\lambda<254 \mathrm{nm} ?\)

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