Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
By relativistic energy, the proton speed is 0.024c.
We need to know about relativistic energy to solve this problem. The rest energy of the object can be determined by
Eo = m₀ . c²
where Eo is rest energy, m₀ is rest mass and c is speed of light (3 x 10⁸ m/s).
The total energy of object can be described as
E = Eo / √(1 - v²/c²)
where E is total energy, v is the object speed.
The kinetic energy is
KE = E - Eo
From the question above, we know that :
mp = 1.6 x 10¯²⁷ kg
me = 9.1 x 10¯³¹ kg
c = 3 x 10⁸ m/s
ve = 0.75c
Find the rest energy of electron
Eo = me . c²
Eo = 9.1 x 10¯³¹ . (3 x 10⁸)²
Eo = 8.19 x 10¯¹⁴ joule
Eo = 8.19 x 10¯¹⁴ / (1.6 x 10¯¹⁹) eV
Eo = 511875 eV
Determine the total energy of electron
E = Eo / √(1 - ve²/c²)
E = 511875 / √(1 - (0.75c)²/c²)
E = 511875 / 0.66
E = 773882.26 eV
Calculate the kinetic energy of electron
KE = E - Eo
KE = 773882.26 - 511875
KE = 262007.26 eV
Find the rest energy of proton
Eo = mp . c²
Eo = 1.6 x 10¯²⁷ . (3 x 10⁸)²
Eo = 1.44 x 10¯¹⁰ joule
Eo = 1.44 x 10¯¹⁰ / (1.6 x 10¯¹⁹) eV
Eo = 900000000 eV
Determine the total energy of proton
E = KE + Eo
E = 262007.26 + 900000000
E = 900262007.3 eV
Find the speed of proton
900262007.3 = 900000000 / √(1 - v²/c²)
√(1 - v²/c²) = 0.9997
1 - v²/c² = 0.9994
v²/c² = 1 - 0.9994
v²/c² = 5.82 x 10¯⁴
v² = 5.82 x 10¯⁴ c²
v = 0.024c
Find more on relativistic energy at: brainly.com/question/23920189
#SPJ4
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.