I believe the answer is Electron
<span>Sometimes when you are training for event you may notice a decrease in improvement in level out of performance. T</span>rue
Answer:
![v = 3.65 * 10^6 m/s](https://tex.z-dn.net/?f=v%20%3D%203.65%20%2A%2010%5E6%20m%2Fs)
Explanation:
Given
de Broglie wavelength = 0.20nm
Required
Determine the speed (v)
The speed is calculated using the following formula;
![L = \frac{h}{mv}](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7Bh%7D%7Bmv%7D)
Where
![L = Wavelength = 0.2nm](https://tex.z-dn.net/?f=L%20%3D%20Wavelength%20%3D%200.2nm)
![h = Planck's\ constant = 6.63 * 10^{-34}Js](https://tex.z-dn.net/?f=h%20%3D%20Planck%27s%5C%20constant%20%3D%206.63%20%2A%2010%5E%7B-34%7DJs)
![m = Mass\ of\ Electron = 9.11 * 10^{-31} kg](https://tex.z-dn.net/?f=m%20%3D%20Mass%5C%20of%5C%20Electron%20%3D%209.11%20%2A%2010%5E%7B-31%7D%20kg)
Substitute these values in the above formula
![0.2nm = \frac{6.63 * 10^{-34}}{9.11 * 10^{-31} * v}](https://tex.z-dn.net/?f=0.2nm%20%3D%20%5Cfrac%7B6.63%20%2A%2010%5E%7B-34%7D%7D%7B9.11%20%2A%2010%5E%7B-31%7D%20%2A%20v%7D)
-----------------------------------------------------
Convert 0.2nm to metre (m)
![0.2nm = 0.2 * 10^{-9}m](https://tex.z-dn.net/?f=0.2nm%20%3D%200.2%20%2A%2010%5E%7B-9%7Dm)
-----------------------------------------------------
![0.2 * 10^{-9} = \frac{6.63 * 10^{-34}}{9.11 * 10^{-31} * v}](https://tex.z-dn.net/?f=0.2%20%2A%2010%5E%7B-9%7D%20%3D%20%5Cfrac%7B6.63%20%2A%2010%5E%7B-34%7D%7D%7B9.11%20%2A%2010%5E%7B-31%7D%20%2A%20v%7D)
Multiply both sides by v
![v * 0.2 * 10^{-9} = \frac{6.63 * 10^{-34}}{9.11 * 10^{-31} * v} * v](https://tex.z-dn.net/?f=v%20%2A%200.2%20%2A%2010%5E%7B-9%7D%20%3D%20%5Cfrac%7B6.63%20%2A%2010%5E%7B-34%7D%7D%7B9.11%20%2A%2010%5E%7B-31%7D%20%20%2A%20v%7D%20%2A%20v)
![v * 0.2 * 10^{-9} = \frac{6.63 * 10^{-34}}{9.11 * 10^{-31} }](https://tex.z-dn.net/?f=v%20%2A%200.2%20%2A%2010%5E%7B-9%7D%20%3D%20%5Cfrac%7B6.63%20%2A%2010%5E%7B-34%7D%7D%7B9.11%20%2A%2010%5E%7B-31%7D%20%7D)
![v * 0.2 * 10^{-9} = \frac{0.73 * 10^{-34}}{10^{-31} }](https://tex.z-dn.net/?f=v%20%2A%200.2%20%2A%2010%5E%7B-9%7D%20%3D%20%5Cfrac%7B0.73%20%2A%2010%5E%7B-34%7D%7D%7B10%5E%7B-31%7D%20%7D)
![v * 0.2 * 10^{-9} = 0.73 * 10^{-34} * 10^{31}](https://tex.z-dn.net/?f=v%20%2A%200.2%20%2A%2010%5E%7B-9%7D%20%3D%200.73%20%2A%2010%5E%7B-34%7D%20%2A%2010%5E%7B31%7D)
Apply law of indices
![v * 0.2 * 10^{-9} = 0.73 * 10^{-34 + 31}](https://tex.z-dn.net/?f=v%20%2A%200.2%20%2A%2010%5E%7B-9%7D%20%3D%200.73%20%2A%2010%5E%7B-34%20%2B%2031%7D)
![v * 0.2 * 10^{-9} = 0.73 * 10^{-3}](https://tex.z-dn.net/?f=v%20%2A%200.2%20%2A%2010%5E%7B-9%7D%20%3D%200.73%20%2A%2010%5E%7B-3%7D)
Divide both sides by ![0.2 * 10^{-9}m](https://tex.z-dn.net/?f=0.2%20%2A%2010%5E%7B-9%7Dm)
![v = \frac{0.73 * 10^{-3} }{ 0.2 * 10^{-9} }](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B0.73%20%2A%2010%5E%7B-3%7D%20%7D%7B%200.2%20%2A%2010%5E%7B-9%7D%20%7D)
![v = \frac{3.65 * 10^{-3} }{10^{-9} }](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B3.65%20%2A%2010%5E%7B-3%7D%20%7D%7B10%5E%7B-9%7D%20%7D)
![v = \frac{3.65 * 10^{-3} }{10^{-9} }](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B3.65%20%2A%2010%5E%7B-3%7D%20%7D%7B10%5E%7B-9%7D%20%7D)
Apply law of indices
![v = 3.65 * 10^{-3} * 10^9](https://tex.z-dn.net/?f=v%20%3D%203.65%20%2A%2010%5E%7B-3%7D%20%2A%2010%5E9)
![v = 3.65 * 10^{-3+9}](https://tex.z-dn.net/?f=v%20%3D%203.65%20%2A%2010%5E%7B-3%2B9%7D)
![v = 3.65 * 10^6](https://tex.z-dn.net/?f=v%20%3D%203.65%20%2A%2010%5E6)
Hence;
The velocity is
![v = 3.65 * 10^6 m/s](https://tex.z-dn.net/?f=v%20%3D%203.65%20%2A%2010%5E6%20m%2Fs)
'D' would do the job ... When you subtract the protons from the mass,
what you have left is neutrons. (The electrons can be ignored. It takes
around 1840 electrons ! to add the mass of a single proton or neutron !)
I don't know it for a fact, but I'd be surprised if the process is really that
simple. I mean, it starts out with knowing the atomic mass, and then
knowing the number of protons in the nucleus. Each of those is a
whole complex problem in itself.
C) is the correct answer.