Asked by Marium
Q)1 : Alpha particles travel through a magnetic field of 0.360T and are deflected in an arc of 0.0820m. Assuming the alpha particle are perpendicular to the field, what is the energy of an individual alpha particle?
Answers
Answered by
bobpursley
calculate the velocity:
Bqv=Force
and force= massalpha*v^2/r
bqv=massv^2/r
solve for v.
Bqv=Force
and force= massalpha*v^2/r
bqv=massv^2/r
solve for v.
Answered by
Marium
do i use the value of q/m ratio??
and thank u =)
and thank u =)
Answered by
jeevakiran
Magnetic field B = 0.36 T
Deflected radius of an arc r = 0.082 m
Charge of alpha particle q = 2e = 3.20435 × 10^-19 C
Mass of alpha particle m = 6.64 × 10^−27 kg
Magnetic Force = Bqv
Centripetal Force = mv²/r
Bqv = mv²/r
v = Bqr/m
v = (0.36)( 3.20435 × 10^-19)( 0.082) / (6.64 × 10^−27)
v = 1.42458 × 10^6 m/s
Energy of alpha particle E = (1/2)mv²
E = (1/2)( 6.64 × 10−27)( 1.42458 × 106)²
E = 4.7296 × 10^−15 J
Energy of alpha particle is 4.7296 × 10^−15 J.
Deflected radius of an arc r = 0.082 m
Charge of alpha particle q = 2e = 3.20435 × 10^-19 C
Mass of alpha particle m = 6.64 × 10^−27 kg
Magnetic Force = Bqv
Centripetal Force = mv²/r
Bqv = mv²/r
v = Bqr/m
v = (0.36)( 3.20435 × 10^-19)( 0.082) / (6.64 × 10^−27)
v = 1.42458 × 10^6 m/s
Energy of alpha particle E = (1/2)mv²
E = (1/2)( 6.64 × 10−27)( 1.42458 × 106)²
E = 4.7296 × 10^−15 J
Energy of alpha particle is 4.7296 × 10^−15 J.
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