CBSE-XI-Physics
05: Newton's Laws of Motion
- #3In a TV picture tube, electrons are ejected from the cathode with negligible speed and they attain a velocity of 5 × 106 m/s in travelling one centimetre. Assuming straight-line motion, find the constant force exerted on the electrons. The mass of an electron is 9.1 × 10-31 kg.Ans : Initial velocity of the electrons is negligible, i.e. u = 0.
Final velocity of the electrons, v = 5 × 106 m/s
Distance travelled by the electrons,
s = 1 cm = 1 × 10-2 m
∴ Acceleration, `` a=\frac{{v}^{2}-{u}^{2}}{2\,\mathrm{\,S \,}}``
`` \Rightarrow a=\frac{{\left(5\times {10}^{5}\right)}^{2}-0}{2\times 1\times {10}^{-3}}=\frac{25\times {10}^{12}}{2\times {10}^{-2}}``
`` \Rightarrow a=12.5\times {10}^{14}\,\mathrm{\,m \,}/{\,\mathrm{\,s \,}}^{2}``
So, force on the electrons, F = ma
⇒ F = 9.1 × 10-31 × 12.5 × 10-14
= 1.1 × 10-15 N
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