Electron deflection

Classical Mechanics Level 2

An electron with a speed of $1.2 \times 10^7 \text{ m/s}$ moves horizontally into a region where a constant vertical force of $4.5\times 10^{-16} \text{ N}$ acts on it. The mass of the electron is $9.11\times 10^{-31}\text{ kg}$ .

Determine the vertical distance the electron is deflected during the time it has moved $30\text{ mm}$ horizontally.

2 \text{ mm}

1.5 \text{ mm}

2.5 \text{ mm}

0.5 \text{ mm}

1 solution

Kaushik Chandra
Jan 5, 2017

So, the velocity of electron, v =1.2×10^7m/s Force, F applied in the 30mm journey of the electron is 4.5×10^-16N and the mass, m of the electron is 9.11×10^-31kg. According to the Newton's 2nd law , F=ma so, a=F/m =4.5×10^-16 /9.11×10^-31 ms^-2 Hence, a =4.94×10^14 ms^-2

Time taken by the electron to go 30mm =distance /speed =0.03/1.2 ×10^7 = 2.5×10^-19 s. [Remember that force acts downward so electron's horizontal velocity is unchanged. Only its vertical motion is affected which we see as deflection in this problem ] Deflection in vertical direction is given by the distance travelled in the vertical direction under acceleration, a and time interval , t. S = 0.5 at^2 = 0.5×4.94.10^14×(2.5×10^-19)^2 =1.5mm.

THANK YOU!

Electron deflection

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