A proton (mass 1.67 × 10 -27 kg) is accelerated along a straight line by 3.6 × 10 15 ms -2 . If the length covered is 3.5 cm and initial speed of proton was 2.4 × 10 7 ms -1 , the gain in kinetic energy is
Text Solution
Verified by ExpertsThe correct answer is:
A
Using v 2 – u 2 = 2as, we get
v = (u 2 + 2aS) 1/2
= (2.4 × 10 7 ) + 2 × 3.6 × 10 15 × 0.035
= 2.88 × 10 7 ms -1
Chang in K.E. = 
m (v 2 – u 2 )
= 
(1.67 × 10 -27 ) (2.88 × 10 7 – 2.4 × 10 7 )
= 2.11 × 10 -13 J
= 1.32 × 10 6 eV = 1.32 MeV
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