In an experiment for positive ray analysis with Thomson method, two identical parabola are obtianed when applied electric fields are `3000` and `2000 V//m`. The particles are singly ionised particles assuming same magnetic field :A. `1 : 3`B. `2 : 4`C. `3 : 1`D. `4 : 2`

In an experiment for positive ray analysis with Thomson method, two id

1.	In an experiment for positive ray analysis with Thomson method, two identical parabola are obtianed when applied electric fields are `3000` and `2000 V//m`. The particles are singly ionised particles assuming same magnetic field :A. `1 : 3`B. `2 : 4`C. `3 : 1`D. `4 : 2`
Answer» Correct Answer - A For same magnetic field `(y^(2))/(x) prop (1)/(E) ((e)/(m))` For singly ionised particle, `(y^(2))/(x) prop (1)/(E_(1)) (e)/(m_(1))` Fro doubly ionised particle, `(y^(2))/(x) prop (1)/(E_(2)) (2e)/(m_(2))` Since the parabolas for both the particles are identical, `(1)/(E_(1)) (e)/(m_(1)) = (1)/(E_(2)) (2e)/(m_(2))` So `(m_(1))/(m_(2)) = (E_(2))/(2E_(1)) = (2000)/(2 xx 3000) = (1)/(3)`

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In an experiment for positive ray analysis with Thomson method, two identical parabola are obtianed when applied electric fields are `3000` and `2000 V//m`. The particles are singly ionised particles assuming same magnetic field :A. `1 : 3`B. `2 : 4`C. `3 : 1`D. `4 : 2`

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