Question

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

A) 3MeV electron
B) 10MeV alpha
C) 0.1 MeV auger
D) 400keV proton

Answer 1

Answer: 3MeV electron

Step-by-step 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 =
`(1)/(2)*{m_e}*{v_(e) ^(2)}`=3MeV

`v_(e) ^(2)=(2*3*1.6*10^(-19)*10^(6)  )/(9.1* 10^(-31) )`=1.05\times10^{18}

`v_e=\sqrt{1.05*10^(18) } = 1.025*10^(9)(m)/(s)`

(b) K.E. Energy of alpha particle =
`(1)/(2)*{m_\alpha}*{v_(\alpha) ^(2)}`=10MeV

`v_(\alpha) ^(2)= (2*10*1.6*10^(-19)*10^(6)  )/(4*9.1* 10^(-31) )`=0.88\times10^{18}

`v_\alpha=\sqrt{0.88*10^(18) } =.94*10^(9)(m)/(s)`

(c) K.E. Energy of auger particle =
`(1)/(2)*{m_a}*{v_(a) ^(2)}`=0.1MeV

`v_(a) ^(2)=(2*0.1*1.6*10^(-19)*10^(6)  )/(9.1* 10^(-31) )`=0.035\times10^{18}

`v_a=\sqrt{0.035*10^(18) } =.19*10^(9)(m)/(s)`

(d) K.E. Energy of proton particle =
`(1)/(2)*{m_p}*{v_(p) ^(2)}`=400keV

`v_(p) ^(2)=(2*400*1.6*10^(-19)*10^(3)  )/(1.67* 10^(-27) )`=0.766\times10^{14}

`v_p=\sqrt{0.766*10^(14) } =0.875*10^(7)[tex](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.