Question

Protons can be accelerated in particle accelerators. Calculate the wavelength (in Å) of such accelerated proton moving at 2.85 10⁸ ms⁻¹ ( the mass of proton is 1.673 10⁻²⁷ Kg).

Answer 1

This question uses the concepts of Special Relativity and De-Broglie's Hypothesis.


We are given that a proton is accelerated to extremely high speeds in a Particle Accelerator. 

We are given the following data:

$\text{Rest Mass } = m_{\circ} = 1.673 \times 10^{-27} \, \, kg \\ \\ \text{Velocity } = v = 2.85 \times 10^8 \, \, m/s$

Now, the speed at which the proton is very close to the speed of light. This is what we call Relativistic Speeds. 

Now, Special Relativity's Relativistic Mass Equation is:

$m = \frac{m_{\circ}}{\sqrt{1-\frac{v^2}{c^2}}}$

Here,

m = Relativistic Mass
c = Speed of light = $3\times 10^8$  m/s


Thus, we see that the mass to be considered at extremely high speeds is greater than the rest mass.


Let us calculate the relativistic mass of the proton:

$m = \frac{m_{\circ}}{\sqrt{1-\frac{v^2}{c^2}}} \\ \\ \\ \implies m = \frac{m_{\circ}}{\sqrt{1-\left(\frac{2.85\times 10^8}{3\times 10^8}\right)^2}} \\ \\ \\ \implies m = \frac{m_{\circ}}{\sqrt{1-(0.95)^2}} \\ \\ \\ \implies m = \frac{m_{\circ}}{\sqrt{1-0.9025}} \\ \\ \\ \implies m = \frac{m_{\circ}}{\sqrt{0.0975}} \\ \\ \\ \implies m \approx \frac{m_{\circ}}{0.3122} \\ \\ \\ \implies m \approx m_{\circ} \times 3.202 \\ \\ \\ \implies m \approx 3.202\times 1.673 \times 10^{-27}$


$\implies m \approx 5.358 \times 10^{-27} \, \, kg$

Thus, now we have the relativistic mass of the proton. Now, we have to find the wavelength, for which we use the De-Broglie Hypothesis. 

De-Broglie stated that all matter particles also behave like waves, and he gave a mathematical formulation as:

$\lambda = \frac{h}{p}$

Here,

$\lambda$ = Wavelength of matter particle
$h$ = Planck's Constant = $6.626 \times 10^{-34} \, \, J \, s$
$p$ = Momentum of Matter Particle = $m\times v$

Now, we can find the wavelength of the relativistically accelerated proton:

$\lambda = \frac{h}{p} \\ \\ \\ \implies \lambda = \frac{h}{mv} \\ \\ \\ \implies \lambda = \frac{6.626 \times 10^{-34}}{5.358 \times 10^{-27} \times 2.85 \times 10^8} \\ \\ \\ \implies \lambda = \frac{6.626 \times 10^{-34}}{1.527 \times 10^{-18}} \\ \\ \\ \implies \lambda \approx 4.34 \times 10^{-16}\, \, m \\ \\ \\ \implies \boxed{\lambda = 0.00000434 \, \, \AA}$



Thus, the wavelength of the proton is $0.00000434 \, \, \AA$ , which can be more conveniently written as $4.34 \times 10^{-16} \, \,metres$


Hope it helps
Purva
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