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]
Answer (integer)
543
Solution
V = 12 kV<br><br>Number of revolution = n<br><br>$n[2 \times {q_P} \times V] = {1 \over 2}{m_P} \times v_P^2$<br><br>$n[2 \times 1.6 \times {10^{ - 19}} \times 12 \times {10^3}]$<br><br>$$ = {1 \over 2} \times 1.67 \times {10^{ - 27}} \times {\left[ {{{3 \times {{10}^8}} \over 6}} \right]^2}$$<br><br>n(38.4 $\times$ 10<sup>$-$16</sup>) = 0.2087 $\times$ 10<sup>$-$11</sup><br><br>n = 543.4
About this question
Subject: Physics · Chapter: Electromagnetic Induction · Topic: Self and Mutual Inductance
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