Question

An electron and proton are detected in cosmic ray experiment the first with Kinetic energy 10 Kev and
the second with 100 Kev. Which is faster, the electron or proton?
me = 9.1 X10^-31kg mp = 1.67 X 10^-27 kg​

Answer 1

Answer:

Electron is faster.

Explanation:

K = mv²/2

for electron,

K = mv²/2

v²=2K/m

v= √(2K/m)

 = √(2 x 10 x 10³/9.11 x 10⁻³¹)

 = 1.482 x 10¹⁷

for proton

v= √(2 x 100 x 10³/ 1.67 x 10⁻²⁷)

 = 1.094 x 10¹⁶

v for electron is greater than v for proton.

Hence, electron is faster.

Answer 2

Answer:

We know that,

Kinetic Energy = 1/2 mv²

According to the question,

• Kinetic Energy of Proton = 100 Kev
• Kinetic Energy of Electron = 10 Kev

Case 1: Velocity of Proton:

$\implies \dfrac{1}{2} m_pv_p^2 = 100\:\: Kev\\\\\implies 100\:\:Kev = 100 \times 1.6 \times 10^{-16}\:\: J\\\\\implies \dfrac{1}{2} m_pv_p^2 = 1.6 \times 10^{-14}\:\: J\\\\\text{Substituting the mass of proton and transposing we get,}\\\\\implies \dfrac{1}{2} \times 1.67 \times 10^{-27} \times v_p^2 = 1.6 \times 10^{-14}\:\: J\\\\\implies v_p^2 = 1.6 \times 10^{-14}\:\: J \times 2 \times \dfrac{1}{ 1.67 \times 10^{-27}}$

$\implies v_p = \sqrt{1.6 \times 10^{-14} \times 2 \times \dfrac{1}{1.67 \times 10^{-27}}}\\\\\implies v_p = 4.38 \times 10^6\:\: m/s$

Case 2: Velocity of Electron:

$\implies \dfrac{1}{2}m_ev_e^2 = 10\:\:Kev\\\\\implies \dfrac{1}{2}m_ev_e^2 = 10 \times 1.6 \times 10^{-16}\:\:J\\\\\implies \dfrac{1}{2}m_ev_e^2 = 1.6 \times 10^{-15}\:\:J\\\\\implies \dfrac{1}{2} \times 9.1 \times 10^{-31} \times v_e^2 = 1.6 \times 10^{-15}\:\:J\\\\\implies v_e^2 = 1.6 \times 10^{-15} \times 2 \times \dfrac{1}{9.1 \times 10^{-31}}\\\\\implies v_e = \sqrt{1.6 \times 10^{-15} \times 2 \times \dfrac{1}{9.1 \times 10^{-31}}}\\\\\\\implies v_e = 5.93 \times 10^7 \;\:m/s$

From the above velocities, we can see that velocity of electron is greater than the velocity of proton.

Therefore electron having 10 Kev is faster than a proton having 100 Kev.