A cathode ray tube contains a pair of parallel metal plates `1.0 cm` apart and `3.0 cm` long. A narrow horizontal beam of electron with a velocity of `3 xx 10^(7) ms^(-1)` is passed down the tube midway between the two plates. When a potential difference of `550 V` is maintained across the plates, it is found that the electron beam is so deflected that it just strikes the end of one of the plates. then the specific change of an electron (that is, the ratio of its charge to mass) in `C//kg` is :

Options

A. `3.6 xx 10^(-14)C//kg`
B. `1.8 xx 10^(-11) C//kg`
C. `3.6 xx 10^(-12)C//kg`
D. `1.8 xx 10^(-9)C//kg`

Answer

B

Explanation

Here `y = (1)/(2) cm = 0.5 xx 10^(-2)m`
`l = 3 cm = 3 xx 10^(-2)m`
and `v = 3 xx 10^(7) ms^(-1)`
Potential difference, `V = 550` volts
`l = vt` or `t = (1)/(v)`
`:. Y = (1)/(2) at^(2) = (1)/(2) ((eE)/(m)) (t^(2))/(v^(2)) = (1)/(2)e ((V)/(dm)) xx (l^(2))/(v^(2))`
`0.5 xx 10^(-2) = (1)/(2) ((e)/(m)) ((550)/(1 xx 10^(-2))) xx ((3 xx 10^(2))/(3 xx 10^(7)))`
`(e)/(m) = (2 xx 0.5 xx 10^(-2) xx 1 xx 10^(-2) xx 9 xx 10^(14))/(550 xx (3 xx 10^(-2))^(2))`
`= 1.8 xx 10^(11) C//kg`