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QJEEM21Phd09b: JEE Main 2021 Physics - Class 12 JEEM21Phd09b Solution: If the maximum value of accelerating potential provided by a…

If the maximum value of accelerating potential provided by a ratio frequency oscillator is 12 kV. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is ...............

[mp = 1.67 ×\times 10 −- 27 kg, e = 1.6 ×\times 10 −- 19C, Speed of light = 3 ×\times 108 m/s]

Step-by-Step Verified Solution

V = 12 kV

Number of revolution = n

n[2×qP×V]=12mP×vP2n[2 \times {q_P} \times V] = {1 \over 2}{m_P} \times v_P^2

n[2×1.6×10−19×12×103]n[2 \times 1.6 \times {10^{ - 19}} \times 12 \times {10^3}]

=12×1.67×10−27×[3×1086]2= {1 \over 2} \times 1.67 \times {10^{ - 27}} \times {\left[ {{{3 \times {{10}^8}} \over 6}} \right]^2}

n(38.4 ×\times 10 −- 16) = 0.2087 ×\times 10 −- 11

n = 543.4

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