JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 20)
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]
[mp = 1.67 $$\times$$ 10$$-$$27 kg, e = 1.6 $$\times$$ 10$$-$$19C, Speed of light = 3 $$\times$$ 108 m/s]
Answer
543
Explanation
V = 12 kV
Number of revolution = n
$$n[2 \times {q_P} \times V] = {1 \over 2}{m_P} \times v_P^2$$
$$n[2 \times 1.6 \times {10^{ - 19}} \times 12 \times {10^3}]$$
$$ = {1 \over 2} \times 1.67 \times {10^{ - 27}} \times {\left[ {{{3 \times {{10}^8}} \over 6}} \right]^2}$$
n(38.4 $$\times$$ 10$$-$$16) = 0.2087 $$\times$$ 10$$-$$11
n = 543.4
Number of revolution = n
$$n[2 \times {q_P} \times V] = {1 \over 2}{m_P} \times v_P^2$$
$$n[2 \times 1.6 \times {10^{ - 19}} \times 12 \times {10^3}]$$
$$ = {1 \over 2} \times 1.67 \times {10^{ - 27}} \times {\left[ {{{3 \times {{10}^8}} \over 6}} \right]^2}$$
n(38.4 $$\times$$ 10$$-$$16) = 0.2087 $$\times$$ 10$$-$$11
n = 543.4
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