To calculate the speed of the Proton we use, de Broglie equation as
[tex]\lambda =\frac{h}{mv}[/tex]
Here, m is mass of proton and its value of [tex]1.6726219 \times 10^{-27} kg[/tex] and h is Plank constant and its value of [tex]6.626\times 10^{-34} Js[/tex].
Given, [tex]\lambda = 156 pm = 156 \times 10^{-12} m[/tex] .
Substituting these values in above equation, we get
[tex]156 \times 10^{-12} m = \frac{6.626\times 10^{-34} Js}{1.6726219 \times 10^{-27} kg\times v} \\\\ v=\frac{3.96\times 10^{-7} }{156 \times 10^{-12} } = 0.02538 \times 10^{5} \ m/s\\\\ v = 2538 \ m/s[/tex]
Thus, the speed of proton is 2538 m/s.
The speed of the proton in the linear accelerator with the given de broglie wavelength is 2539.67m/s
Given the data in the question;
De Broglie wavelength of the proton; [tex]\lambda = 156pm = 1.56 * 10^{-10}m[/tex]
Speed of the proton; [tex]v = \ ?[/tex]
The de Broglie wavelength of any particle shows the length scale at which wave-like properties are important for such particle. It is expressed as:
[tex]\lambda = \frac{h}{mv}[/tex]
Where:
We substitute our values into the equation
[tex]1.56*10^{-10}m = \frac{6.626*10^{-34}J.s}{(1.6726*10^{-27}kg)\ *\ v}\\\\v = \frac{6.626*10^{-34}J.s}{(1.6726*10^{-27}kg)*(156*10^{-10}m) }\\\\v = \frac{6.626*10^{-34}J.s}{2.609*10^{-37}kg.m}\\\\v = \frac{6.626*10^{-34}kg.m^2/s^{2}.s}{2.609*10^{-37}kg.m}\\\\v = \frac{6.626*10^{-34}kg.m^2/s}{2.609*10^{-37}kg.m}\\\\v = 2539.67 m/s[/tex]
Therefore, the speed of the proton in the linear accelerator with the given de broglie wavelength is 2539.67m/s.
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