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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 脳 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 脳 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

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