A5.If the de Broglie wavelength is the same , the highest velocity is __ 1. (}_{1)^0e 2. (}_{1)^1P 3. (}_{0)^1n 4. (}_{2)^4He

To determine which particle has the highest velocity given the same de Broglie wavelength, we need to consider the relationship between momentum and velocity. The de Broglie wavelength \(\lambda\) is given by:<br /><br />\[ \lambda = \frac{h}{p} \]<br /><br />where \(h\) is Planck's constant and \(p\) is the momentum of the particle. Since the wavelength is the same for all particles, their momenta must be equal. Momentum \(p\) is related to velocity \(v\) by:<br /><br />\[ p = mv \]<br /><br />where \(m\) is the mass of the particle. Therefore, for the same momentum, the velocity is inversely proportional to the mass:<br /><br />\[ v = \frac{p}{m} \]<br /><br />Given the particles:<br />1. \({}_{1}^{0}e\) (electron)<br />2. \({}_{1}^{1}P\) (proton)<br />3. \({}_{0}^{1}n\) (neutron)<br />4. \({}_{2}^{4}He\) (alpha particle or helium nucleus)<br /><br />The masses of these particles are approximately:<br />- Electron: \(9.109 \times 10^{-31} \, \text{kg}\)<br />- Proton: \(1.673 \times 10^{-27} \, \text{kg}\)<br />- Neutron: \(1.675 \times 10^{-27} \, \text{kg}\)<br />- Alpha particle: \(6.644 \times 10^{-27} \, \text{kg}\)<br /><br />Since the electron has the smallest mass, it will have the highest velocity for the same de Broglie wavelength.<br /><br />Therefore, the correct answer is:<br />1. \({}_{1}^{0}e\)