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Chapter 7: Problem 51

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

Short Answer

Expert verified
The mass of the particle is approximately \(9.09 \times 10^{-31} \mathrm{~kg}\).

Step by step solution

01

Calculate momentum using de Broglie wavelength formula

The de Broglie wavelength formula is given by: \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is the Planck's constant, and p is the momentum of the particle. We can rearrange the formula to get the momentum: \( p = \frac{h}{\lambda} \) Given the wavelength as \(1.5 \times 10^{-15}\mathrm{~m}\), Momentum, \( p = \frac{6.626 \times 10^{-34}\mathrm{~J \ s}}{1.5 \times 10^{-15}\mathrm{~m}} \)
02

Calculate the relativistic momentum

Now we know the momentum of the particle. Since we are dealing with a particle moving close to the speed of light, we will use the relativistic momentum formula: \( p = m \cdot v \cdot \frac{1}{\sqrt{1 - (\frac{v^2}{c^2})}} \), where m is the mass of the particle, v is the velocity, and c is the speed of light. The velocity is given as 90% of the speed of light: \( v = 0.9 \cdot c = 0.9 \times 3 \times 10^{8}\mathrm{~m/s} \) Now, we can plug in the values and solve for the mass: \( m = \frac{p}{v \cdot \frac{1}{\sqrt{1 - (\frac{v^2}{c^2})}}} \)
03

Simplify and solve for the mass

Plug in the values for p, v, and c into the formula above: \( m = \frac{\frac{6.626 \times 10^{-34}\mathrm{~J \ s}}{1.5 \times 10^{-15}\mathrm{~m}}}{(0.9 \times 3 \times 10^{8}\mathrm{~m/s}) \cdot \frac{1}{\sqrt{1 - (\frac{(0.9 \times 3 \times 10^{8}\mathrm{~m/s})^2}{(3 \times 10^{8}\mathrm{~m/s})^2})}}} \) Solve for m: \( m \approx 9.09 \times 10^{-31} \mathrm{~kg} \) Therefore, the mass of the particle is approximately \(9.09 \times 10^{-31} \mathrm{~kg}\).

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Most popular questions from this chapter

Cesium was discovered in natural mineral waters in 1860 by R. W. Bunsen and G. R. Kirchhoff using the spectroscope they invented in \(1859 .\) The name came from the Latin caesius ("sky blue") because of the prominent blue line observed for this element at \(455.5 \mathrm{~nm} .\) Calculate the frequency and energy of a photon of this light.

Which of the following electron configurations correspond to an excited state? Identify the atoms and write the ground-state electron configuration where appropriate. a. \(1 s^{2} 2 s^{2} 3 p^{1}\) b. \(1 s^{2} 2 s^{2} 2 p^{6}\) c. \(1 s^{2} 2 s^{2} 2 p^{4} 3 s^{1}\) d. \([\mathrm{Ar}] 4 s^{2} 3 d^{5} 4 p^{1}\) How many unpaired electrons are present in each of these species?

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{~s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

Using only the periodic table inside the front cover of the text, write the expected ground-state electron configurations for a. the third element in Group \(5 \mathrm{~A}\). b. element number 116 . c. an element with three unpaired \(5 d\) electrons. d. the halogen with electrons in the \(6 p\) atomic orbitals.

Write the expected ground-state electron configuration for the following. a. the element with one unpaired \(5 p\) electron that forms a covalent with compound fluorine b. the (as yet undiscovered) alkaline earth metal after radium c. the noble gas with electrons occupying \(4 f\) orbitals d. the first-row transition metal with the most unpaired electrons
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