Respuesta :
Answer: 3MeV electron
Explanation:
m_e={9.1\times 10^{-31} m_α=4\times m_e m_a={9.1\times 10^{-31}
m_p=1.67\times 10^{-27}
(a) K.E. Energy of electron =[tex]\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}[/tex]=3MeV
[tex]v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }[/tex]=1.05\times10^{18}
[tex]v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}[/tex]
(b) K.E. Energy of alpha particle =[tex]\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}[/tex]=10MeV
[tex]v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6} }{4\times9.1\times 10^{-31} }[/tex]=0.88\times10^{18}
[tex]v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}[/tex]
(c) K.E. Energy of auger particle =[tex]\frac{1}{2}\times{m_a}\times{v_{a} ^{2}}[/tex]=0.1MeV
[tex]v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }[/tex]=0.035\times10^{18}
[tex]v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}[/tex]
(d) K.E. Energy of proton particle =[tex]\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}[/tex]=400keV
[tex]v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3} }{1.67\times 10^{-27} }[/tex]=0.766\times10^{14}
[tex]v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}[/tex]
from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.
