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Chapter 7: Problem 15

Cobalt- 60 is a radioactive isotope used to treat cancers. A gamma ray emitted by this isotope has an energy of \(1.33 \mathrm{MeV}\) (million electron volts; \(1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\) ). What are the frequency (in \(\mathrm{Hz}\) ) and the wavelength (in \(\mathrm{m}\) ) of this gamma ray?

Short Answer

Expert verified
Frequency: \(3.211 \times 10^{20} \text{~Hz}\), Wavelength: \(9.34 \times 10^{-13} \text{~m}\)

Step by step solution

01

- Convert Energy from MeV to Joules

Energy of the gamma ray is given as 1.33 MeV. Convert it to Joules using the conversion factor: 1 MeV = 1.602 × 10^{-13} J. Thus, energy in Joules is calculated as follows: \[ E = 1.33 \times 1.602 \times 10^{-13} \text{~J} = 2.12866 \times 10^{-13} \text{~J} \]
02

- Calculate the Frequency

Use the energy-frequency relation for photons: \[ E = h u \] Where \(E\) is the energy, \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{~Js})\), and \( u \) is the frequency. Solve for frequency \(u\): \[ u = \frac{E}{h} = \frac{2.12866 \times 10^{-13} \text{~J}}{6.626 \times 10^{-34} \text{~Js}} \ u = 3.211 \times 10^{20} \text{~Hz} \]
03

- Calculate the Wavelength

Use the wave equation relating wave speed, frequency, and wavelength: \[ c = u \times \text{wavelength} \] Here, \(c\) is the speed of light \(( 3 \times 10^8 \text{~m/s})\). Solve for the wavelength: \[ \text{wavelength} = \frac{c}{u} = \frac{3 \times 10^8 \text{~m/s}}{3.211 \times 10^{20} \text{~Hz}} \ \text{wavelength} = 9.34 \times 10^{-13} \text{~m} \]

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Key Concepts

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

radioactive isotope
A radioactive isotope is an atom with an unstable nucleus that loses energy by emitting radiation during its decay to a stable form. Cobalt-60 is such an isotope and is commonly used in medical treatments, such as cancer therapy. Cobalt-60 decays by beta decay and emits gamma rays as a byproduct. These gamma rays are what makes Cobalt-60 effective in destroying cancerous cells.
energy to Joules conversion
To find the frequency and wavelength of a gamma ray emitted by Cobalt-60, we need to convert its energy from million electron volts (MeV) to Joules (J). The energy given is 1.33 MeV, and the conversion factor is 1 MeV = 1.602 × 10^{-13} J. So, \[ E = 1.33 \times 1.602 \times 10^{-13} \text{~J} = 2.12866 \times 10^{-13} \text{~J} \] This conversion is crucial for solving further calculations involving energy, frequency, and wavelength.
frequency calculation
Once the energy is converted to Joules, we can use the formula for energy-frequency relation given by \[ E = h u \] where \( E \) is energy, \( h \) is Planck's constant \ (6.626 \times 10^{-34} \text{~Js}) \, and \( u \) is the frequency. Rearranging the formula to solve for the frequency \( u \), we have: \[ u = \frac{E}{h} = \frac{2.12866 \times 10^{-13} \text{~J}}{6.626 \times 10^{-34} \text{~Js}} \ u = 3.211 \times 10^{20} \text{~Hz} \] This gives us the frequency of the gamma ray in hertz (Hz).
wavelength calculation
The relationship between wavelength, frequency, and the speed of light \( c \) is given by: \[ c = u \times \text{wavelength} \] Where \( c \) is the speed of light (\(3 \times 10^8 \text{~m/s})\). Rearranging this equation to solve for wavelength, we get: \[ \text{wavelength} = \frac{c}{u} = \frac{3 \times 10^8 \text{~m/s}}{3.211 \times 10^{20} \text{~Hz}} \] \[ \text{wavelength} = 9.34 \times 10^{-13} \text{~m} \ \] This calculation tells us the wavelength of the gamma ray emitted by Cobalt-60 in meters.

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Most popular questions from this chapter

What key assumption of Bohr's model would a "Solar System" model of the atom violate? What was the theoretical basis for this assumption?

Define each of the following wave phenomena, and give an example of where each occurs: (a) refraction; (b) diffraction; (c) dispersion; (d) interference.

How many orbitals in an atom can have each of the following designations: (a) \(5 f\) (b) \(4 p_{i}\) (c) \(5 d ;\) (d) \(n=2 ?\)

In the course of developing his model, Bohr arrived at the following formula for the radius of the electron's orbit: \(r_{n}=\) \(n^{2} h^{2} \varepsilon_{0} / \pi m_{0} e^{2},\) where \(m_{c}\) is the electron's mass, \(e\) is its charge, and \(\varepsilon_{0}\) is a constant related to charge attraction in a vacuum. Given that \(m_{\mathrm{z}}=9.109 \times 10^{-31} \mathrm{~kg}, e=1.602 \times 10^{-19} \mathrm{C},\) and \(\varepsilon_{0}=8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{J} \cdot \mathrm{m}\) calculate the following: (a) The radius of the first \((n=1)\) orbit in the \(\mathrm{H}\) atom (b) The radius of the tenth \((n=10)\) orbit in the \(\mathrm{H}\) atom

A ground-state \(\mathrm{H}\) atom absorbs a photon of wavelength \(94.91 \mathrm{nm},\) and its electron attains a higher energy level. The atom then emits two photons: one of wavelength \(1281 \mathrm{nm}\) to reach an intermediate energy level, and a second to return to the ground state. (a) What higher level did the electron reach? (b) What intermediate level did the electron reach? (c) What was the wavelength of the second photon emitted?
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