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 $= 9.11 \times 10^{-31} \; kg$,proton mass $= 1.67 \times 10^{-27} \; kg$,$1 \; eV = 1.60 \times 10^{-19} \; J$)

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The electron is faster. The ratio of their speeds is $13.54:1$.
Given:
Mass of the electron,$m_e = 9.11 \times 10^{-31} \; kg$
Mass of the proton,$m_p = 1.67 \times 10^{-27} \; kg$
Kinetic energy of the electron,$E_{Ke} = 10 \; keV = 10^4 \; eV = 1.60 \times 10^{-15} \; J$
Kinetic energy of the proton,$E_{Kp} = 100 \; keV = 10^5 \; eV = 1.60 \times 10^{-14} \; J$
Using the formula for kinetic energy $E_K = \frac{1}{2}mv^2$,the velocity is $v = \sqrt{\frac{2E_K}{m}}$.
For the electron:
$v_e = \sqrt{\frac{2 \times 1.60 \times 10^{-15}}{9.11 \times 10^{-31}}} \approx 5.93 \times 10^7 \; m/s$
For the proton:
$v_p = \sqrt{\frac{2 \times 1.60 \times 10^{-14}}{1.67 \times 10^{-27}}} \approx 4.38 \times 10^6 \; m/s$
Comparing the velocities,$v_e > v_p$,so the electron is faster.
The ratio of their speeds is $\frac{v_e}{v_p} = \frac{5.93 \times 10^7}{4.38 \times 10^6} \approx 13.54$.

Explore More

Similar Questions

Calculate the number of photons entering the eye of a duck per second,corresponding to the minimum intensity of white light $(\sim 10^{-10} \ W/m^2)$. Take the area of the duck's eye as $0.4 \ cm^2$ and the average frequency of white light as $6 \times 10^{14} \ Hz$.

Our eye can detect $5 \times 10^4$ photons per second per square meter of green light $(\lambda = 5000 \ \mathring{A})$. If the ear can detect $10^{-13} \ W/m^2$,how many times more sensitive is the eye compared to the ear?

The number of photons emitted from a photocell by a source placed at a distance of $0.5 \ m$ is how many times the number of photons emitted by the same source placed at a distance of $1 \ m$?

If the mean wavelength of light radiated by a $100 \ W$ lamp is $5000 \ \mathring{A}$,then the number of photons radiated per second is:

What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength $5200 \, \mathring{A}$? (Given: $h = 6.63 \times 10^{-34} \, J \cdot s$, $m_e = 9.1 \times 10^{-31} \, kg$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo