In the question number 34, the kinetic energy (in MeV) of the proton beam produced by the accelerator is (radius of dees = 60 cm)

Question

In the question number 34, the kinetic energy (in MeV) of the proton beam produced by the accelerator is (radius of dees = 60 cm) (A) 5 (B) 6.5 (C) 10.6 (D) 12.6

Tardigrade Admin · Accepted Answer

Final velocity of proton is v=r ×2π upsilon=0.6×2×3.14×12×106 =4.5×107 m s-1 ∴ K=(1/2) mv2=(1/2)×(1.67×10-27×(4.5×107)2/1.6×10-13) =10.6 MeV

Q. In the question number $34$ , the kinetic energy (in $M e V$ ) of the proton beam produced by the accelerator is (radius of dees $= 60 c m)$

$5$

$6.5$

$10.6$

$12.6$

Final velocity of proton is
$v = r \times 2 π υ = 0.6 \times 2 \times 3.14 \times 12 \times 1 0^{6}$
$= 4.5 \times 1 0^{7} m s^{- 1}$
$∴ K = \frac{1}{2} m v^{2} = \frac{1}{2} \times \frac{1.67 \times 1 0 ^{- 27} \times ( 4.5 \times 1 0 ^{7} ) ^{2}}{1.6 \times 1 0 ^{- 13}}$
$= 10.6 M e V$

Q. In the question number 34, the kinetic energy (in MeV) of the proton beam produced by the accelerator is (radius of dees =60cm)

5

6.5

10.6

12.6

Final velocity of proton is v=r×2πυ=0.6×2×3.14×12×106 =4.5×107ms−1 ∴K=21​mv2=21​×1.6×10−131.67×10−27×(4.5×107)2​ =10.6MeV

Q. In the question number $34$ , the kinetic energy (in $M e V$ ) of the proton beam produced by the accelerator is (radius of dees $= 60 c m)$

$5$

$6.5$

$10.6$

$12.6$

Final velocity of proton is
$v = r \times 2 π υ = 0.6 \times 2 \times 3.14 \times 12 \times 1 0^{6}$
$= 4.5 \times 1 0^{7} m s^{- 1}$
$∴ K = \frac{1}{2} m v^{2} = \frac{1}{2} \times \frac{1.67 \times 1 0 ^{- 27} \times ( 4.5 \times 1 0 ^{7} ) ^{2}}{1.6 \times 1 0 ^{- 13}}$
$= 10.6 M e V$