**Question 6.12:**

^{–31}kg, proton mass = 1.67 × 10

^{–27}kg, 1 eV = 1.60 × 10

^{–19}J).

__Solution__:Electron is faster; Ratio of speeds is 13.54 : 1

*m*

_{e}= 9.11 × 10

^{–31}kg

Mass of the proton,

*m*

_{p}= 1.67 × 10

^{– 27}kg

Kinetic energy of the electron,

*E*

_{Ke}= 10 keV = 10

^{4}eV

= 10

^{4}× 1.60 × 10

^{–19}

= 1.60 × 10

^{–15}J

Kinetic energy of the proton,

*E*

_{Kp}= 100 keV = 10

^{5}eV = 1.60 × 10

^{–14}J

For the velocity of an electron v

_{e}, its kinetic energy is given by the relation:

E

_{Ke}= (1/2) mv

_{e}

^{2}

∴ v

_{e}= (2E

_{Ke}/ m)

^{1/2}

= (2 X 1.60 X 10

^{-15}/ 9.11 X 10

^{-31})

^{1/2}= 5.93 X 10

^{7}m/s

For the velocity of a proton v

_{p}, its kinetic energy is given by the relation:

_{}

E

_{Kp}= (1/2)mv

_{p}

^{2}

v

_{p}= (2 X 1.6 X 10

^{-14}/ 1.67 X 10

^{-27})

^{1/2}= 4.38 X 10

^{6}m/s

Hence, the electron is moving faster than the proton.

The ratio of their speeds

v

_{e}/ v

_{p}= 5.93 X 10

^{7}/ 4.38 X 10

^{6}= 13.54 / 1

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