Question

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

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

Step-by-step solution

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, \( p = \frac{6.626 \times 10^{-34}\mathrm{~J \ s}}{1.5 \times 10^{-15}\mathrm{~m}} \)

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: \( m = \frac{p}{v \cdot \frac{1}{\sqrt{1 - (\frac{v^2}{c^2})}}} \)

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 \(9.09 \times 10^{-31} \mathrm{~kg}\).