Atomic StructureHard
Question
The potential difference between cathode and anode in a cathode ray tube is V. The speed acquired by the electrons is proportional to
Options
A.V
B.$\sqrt{V}$
C.V2
D.$1/\sqrt{V}$
Solution
$e.V. = \frac{1}{2}mv^{2} \Rightarrow v = \sqrt{2 \times \left( \frac{e}{m} \right) \times V}\alpha\sqrt{V}$
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