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In a parabola spectrograph, the velocities of four positive ions P, Q, R and S are v1, v2, v3 and v4 respectively. Then

(A) v1 > v2 > v3 > v4

(B) v1 < v2 < v3 < v4

(C) v1 = v2 = v3 = v4

(D) v1 << v2 > v3 < v4

1 Answer

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Best answer

The correct option is (A) v1 > v2 > v3 > v4.

Explanation:

In Thomson mass spectrograph. All the charged particle moving with difference velocity but same charge to mass ratio lie on the same parabola. and x ∝ (1/d2

Hence ion whose deflection is less velocity will be more. 

For curve x1 < x2 < x3 < x4 

Hence v1 > v2 > v3 > v4

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