Question

Which of the following can penetrate the deepest (Please explain)

Answer 1

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 =$\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}$=3MeV

$v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }$=1.05\times10^{18}

$v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}$

(b) K.E. Energy of alpha particle =$\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}$=10MeV

$v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6} }{4\times9.1\times 10^{-31} }$=0.88\times10^{18}

$v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}$

(c) K.E. Energy of auger particle =$\frac{1}{2}\times{m_a}\times{v_{a} ^{2}}$=0.1MeV

$v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }$=0.035\times10^{18}

$v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}$

(d)  K.E. Energy of proton particle =$\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}$=400keV

$v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3} }{1.67\times 10^{-27} }$=0.766\times10^{14}

$v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}$

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.