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 equal to 9.11 × 10⁻³¹ kg, proton mass = 1.67 × 10⁻²⁷ Kg, 1 eV = 1.60 × 10⁻¹⁹ J)

Answer 1

we know, Kinetic energy , $K.E=\frac{1}{2}mv^2$

or, $v=\sqrt{\frac{2K.E}{m}}$

here it is clear that velocity of particle is directly proportional to square root of kinetic energy and inversely proportional to mass of particle.

now, $v_e=\sqrt{\frac{2K.E_e}{m_e}}$

= √{2 × 10KeV/9.11 × 10^-31kg}

= √{2 × 10 × 10³ × 1.6 × 10^-19/9.11 × 10^-31}

= √{ 32/9.11 × 10^15}

and $v_p=\sqrt{\frac{2K.E_p}{m_p}}$

= √{2 × 100KeV/1.67 × 10^-27kg}

= √{2 × 100 × 10³ × 1.6 × 10^-19/1.67 × 10^-27}

= √{320/1.67 × 10^11}

here it is clear that electron is more faster than proton.

$\frac{v_e}{v_p}$ = √{32/9.11 × 10^15}/√{320/1.67 × 10^11}

= √{1.67/10 × 9.11 × 10⁴}

= √{1.67/91.1 × 10⁴}

= √{16700/91.1}

= 13.53 : 1

Answer 2

Answer:

Electron is faster; Ratio of speeds is 13.54 : 1

Mass of the electron, me = 9.11 × 10–31 kg

Mass of the proton, mp = 1.67 × 10– 27 kg

Kinetic energy of the electron, EKe = 10 keV = 104 eV

= 104 × 1.60 × 10–19

= 1.60 × 10–15 J

Kinetic energy of the proton, EKp = 100 keV = 105 eV = 1.60 × 10–14 J

For the velocity of an electron ve, its kinetic energy is given by the relation:

EKe = (1/2) mve2

∴ ve = (2EKe / m)1/2

= (2 × 1.60 × 10-15 / 9.11 × 10-31)1/2 = 5.93 × 107 m/s

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

EKp = (1/2)mvp2

vp = (2 × 1.6 × 10-14 / 1.67 × 10-27 )1/2 = 4.38 × 106 m/s

Hence, the electron is moving faster than the proton.

The ratio of their speeds

ve / vp = 5.93 × 107 / 4.38 × 106 = 13.54 : 1