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A cathode ray tube contains a pair of parallel metal plates 2.0cm apart and 3.5cm long. A narrow horizontal beam of electrons with velocity of 3 \times 10^7 \mathrm{~ms}^{-1} is poured down the tube midway between the two plates. When the potential difference of 670 \mathrm{~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. Find the specific charge of an electron.

Option: 1

2.2 \times 10^{11} \mathrm{~kg}^{-1}


Option: 2

3.1 \times 10^{-8} \mathrm{~kg}^{-1}


Option: 3

2.9 \times 10^{7} \mathrm{~kg}^{-1}


Option: 4

4.1 \times 10^{-6} \mathrm{~kg}^{-1}


Answers (1)

best_answer

\begin{aligned} & y=\frac{1}{2} \mathrm{~cm}=0.5 \times 10^{-2} \mathrm{~m} \\ & l=3.5 \times 10^{-2} \mathrm{~m} \\ & v=3 \times 10^7 \mathrm{~ms}^{-1} \quad \mathrm{~V}=670 \mathrm{~V} \\ & d=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m} \end{aligned}

Time to cross the tube,

t=\frac{l}{v}=\frac{3.5 \times 10^{-2}}{3 \times 10^7}=1.16 \times 10^{-9} \mathrm{~s}

\begin{aligned} y & =\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{E_e}{m}\right) t^2 \\ & =\frac{1}{2} \frac{(v / d) e}{m} t^2 \end{aligned}

\frac{e}{m}=\frac{2 y d}{V t^2}=\frac{2 \times(0.5 \times 10^-^2) \times 2 \times 10^-^2 \mathrm{~m}}{670 \times\left(1.16 \times 10^{-9}\right)}

      \begin{aligned} & =\frac{2 \times 10^{-4}}{670 \times 1.3456 \times 10^{-18}} \\ & =0.0022 \times 10^{-4+18}=0.0022 \times 10^{14} \\ & =2.2 \times 10^{11} \mathrm{~kg}^{-1} \\ & \end{aligned}

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Riya

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