Q10. What potential difference V is required to accelerate alpha particles from

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Q10. What potential difference V is required to accelerate alpha particles from rest to 10% of the speed of light?
Q. 11. A system consisting entirely of electrons and alpha particles has a net charge of 1.84 X 10^15 C and a net mass of 4.56 X 10^-23 kg. How many electrons and alpha particles are in this system?
Q. 12. An electron travelling to the left with velocity 5 X 10^5 m/sec, and making an angle of 30 degrees with the horizontal, moves into a magnetic field of 1.0 T directed toward the observer. What is the magnitude and direction of the magnetic force on the electron when it first enters the magnetic field region? What is the radius of the helical orbit it executes? A typical 12V car battery can deliver 7.5 x 10^5 C of charge. If the energy supplied by the barrery could be converted entirely to energy, what speed would it give to a 1200 kg car? Q10. What potential difference V is required to accelerate alpha particles from rest to 10% of the speed of light?
Q. 11. A system consisting entirely of electrons and alpha particles has a net charge of 1.84 X 10^15 C and a net mass of 4.56 X 10^-23 kg. How many electrons and alpha particles are in this system?
Q. 12. An electron travelling to the left with velocity 5 X 10^5 m/sec, and making an angle of 30 degrees with the horizontal, moves into a magnetic field of 1.0 T directed toward the observer. What is the magnitude and direction of the magnetic force on the electron when it first enters the magnetic field region? What is the radius of the helical orbit it executes? A typical 12V car battery can deliver 7.5 x 10^5 C of charge. If the energy supplied by the barrery could be converted entirely to energy, what speed would it give to a 1200 kg car?

Explanation / Answer

A)2e*V=1/2*4m*0.01c^2

so V=9000kVolts

B)e=X alpha=Y

-1.6*10^-19X+2*1.6*10^-19Y=1.84*10^-15

so 2Y-X=1.15*10^4

and 1.6*10^-27X+4*1.6*10^-27Y=4.56*10^-23

so 4Y-X=2.85*10^4

so Y=8500 and X=5500

C)r=mvsin(30)/qB=1.6*10^-27*5*10^5*0.5/(1.6*10^-19*1)=2.5mm

D)7.5*10^5*12=1/2*1200*v^2

So v=122.47m/s

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