JEE MAIN - Physics (2021 - 27th August Morning Shift - No. 13)
Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :
lighter ion will be deflected less than heavier ion
lighter ion will be deflected more than heavier ion
both ions will be deflected equally
no ion will be deflected
Explanation
$$r = {P \over {qB}} = {{\sqrt {2mk} } \over {qB}}$$
Given they have same kinetic energy
$$r \propto {{\sqrt m } \over q}$$
$${{{r_1}} \over {{r_2}}} = {{\sqrt 4 } \over 2} \times {3 \over {\sqrt {16} }} = {3 \over 4}$$
$${r_2} = {{4{r_1}} \over 3}$$ (r2 is for heavier ion and and r1 is for lighter ion)
_27th_August_Morning_Shift_en_13_1.png)
$$\sin \theta = {d \over R}$$
$$\theta$$ $$\to$$ Deflection
$$\theta \propto {1 \over R}$$
(R $$\to$$ Radius of path)
$$\because$$ R2 > R1 $$\Rightarrow$$ $$\theta$$2 < $$\theta$$1
Given they have same kinetic energy
$$r \propto {{\sqrt m } \over q}$$
$${{{r_1}} \over {{r_2}}} = {{\sqrt 4 } \over 2} \times {3 \over {\sqrt {16} }} = {3 \over 4}$$
$${r_2} = {{4{r_1}} \over 3}$$ (r2 is for heavier ion and and r1 is for lighter ion)
$$\sin \theta = {d \over R}$$
$$\theta$$ $$\to$$ Deflection
$$\theta \propto {1 \over R}$$
(R $$\to$$ Radius of path)
$$\because$$ R2 > R1 $$\Rightarrow$$ $$\theta$$2 < $$\theta$$1
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