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Chapter 2: Problem 55

A particle has a velocity that is \(90 . \%\) of the speed of light. If the wavelength of the particle is \(1.5 \times 10^{-15} \mathrm{m},\) what is the mass of the particle?

Short Answer

Expert verified
The mass of the particle is approximately \(4.283 \times 10^{-28} \mathrm{kg}\).

Step by step solution

01

Calculate the velocity of the particle

Since the particle's velocity is 90% of the speed of light, let's calculate the actual velocity: \[v = 0.90c\] where \(c\) is the speed of light, which is approximately \(3.0 \times 10^8 \mathrm{m/s}\).
02

Rewrite the de Broglie wavelength formula using the relativistic momentum formula

Substitute the expression for relativistic momentum, \(p = \frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}\), into the de Broglie wavelength formula \(\lambda = \frac{h}{p}\): \[ \lambda = \frac{h}{\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}} \]
03

Solve for the mass of the particle

We are given the wavelength, \(\lambda = 1.5 \times 10^{-15} \mathrm{m}\), and the Planck's constant, \(h \approx 6.626 \times 10^{-34} \mathrm{Js}\). We will now solve for the rest mass, \(m_0\), by rearranging the formula: \[ m_0 = \frac{h\sqrt{1-\frac{v^2}{c^2}}}{v\lambda} \] Now, substitute the given values into the formula: \[ m_0 = \frac{(6.626 \times 10^{-34} \mathrm{Js})\sqrt{1-\frac{(0.90 \times 3.0 \times 10^8 \mathrm{m/s})^2}{(3.0 \times 10^8 \mathrm{m/s})^2}}}{(0.90 \times 3.0 \times 10^8 \mathrm{m/s})(1.5 \times 10^{-15} \mathrm{m})} \]
04

Calculate the mass of the particle

Perform the calculation: \[ m_0 \approx 4.283 \times 10^{-28} \mathrm{kg} \] Thus, the mass of the particle is approximately \(4.283 \times 10^{-28} \mathrm{kg}\).

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Most popular questions from this chapter

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and \(589.6 \mathrm{nm}\). What are the frequency and the energy of a photon of light at each of these wavelengths? What are the energies in \(\mathrm{kJ} / \mathrm{mol} ?\)

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