#### Question

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass = 9.11 × 10^{–31} kg, proton mass = 1.67 ×10^{–27} kg, 1 eV = 1.60 × 10^{–19} J).

#### Solution 1

Electron is faster; Ratio of speeds is 13.54 : 1

Mass of the electron, *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 thr velocity of an electron `v_e`, its kinetic energy is given by the relation:

`E_(K_e) = 1/2mv_e^2`

`:.v_e = sqrt((2xxE_(K_e))/m`

`= sqrt((2xx1.60xx10^(-15))/(9.11xx10^(-31))) = 5.93 xx10^7` m/s

For the velocity of a proton `v_p`, its kinetic energy is given by the relation:

`E_(K_p)=1/2mv_p^2`

`v_p = sqrt((2xxE_(K_p))/m`

`:.v_p = sqrt((2xx1.6xx10^(-14))/(1.67xx10^(-27))) = 4.38 xx 10^6 "m/s"`

Hence the electron ismoving faster than the proton.

The ratio of their speeds:

`v_c/v_p = (5.93xx10^7)/(4.38xx10^(6)) = 13.54:1`

#### Solution 2

Here `K_e = 10 KeV` and `K_p = 100 keV`

`m_e = 9.11 xx 10^(-31) kg and m_p = 1.67 xx 10^(-27) kg`

As K = `1/2mv^2` or v = sqrt`((2K)/m)`

Hence `v_e/v_p = sqrt(K_e/K_pxxm_p/m_e) = sqrt((10keV)/100keV) xx (1.67xx10^(-27) kg)/(9.11xx10^(-31)kg)`

= 13.54

`=> v_e = 13.54 v_p`

Thus electron is travelling faster