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 electrons with a velocity of $3 \times 10^{7} {ms}^{- 1}$ 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 charge of an electron (that is, the ratio of its charge to mass) in C/kg is:

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$3 .6 \times 10^{14} C / kg$

$1.8 \times 10^{11} C / kg$

$3 .6 \times 10^{12} C / kg$

$1.8 \times 10^{9} C / kg$

answer is B.

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Detailed Solution

$\begin{array}{l} here y = \frac{1}{2} cm = 0 .5 \times 10^{- 2} m \\ l = 3 cm = 3 \times 10^{- 2} m \\ and v = 3 \times 10^{7} {ms}^{- 1} \\ Potential difference, V = 550 volts \\ l = vt or t = \frac{1}{v} \\ ∴ y = \frac{1}{2} {at}^{2} = \frac{1}{2} (\frac{eE}{m}) \frac{t^{2}}{v^{2}} = \frac{1}{2} e (\frac{V}{dm}) \times \frac{l^{2}}{v^{2}} \\ 0 .5 \times 10^{- 2} = \frac{1}{2} (\frac{e}{m}) (\frac{550}{1 \times 10^{- 2}}) \times (\frac{3 \times 10^{2}}{3 \times 10^{7}}) \\ \frac{e}{m} = \frac{2 \times 0 .5 \times 10^{- 2} \times 1 \times 10^{- 2} \times 9 \times 10^{14}}{550 \times {(3 \times 10^{- 2})}^{2}} = 1 .8 \times 10^{11} C / kg \end{array}$