question_answer

An electron initially at rest falls a distance of \[1.5\,cm\] in a uniform electric field of magnitude \[2\times {{10}^{4}}N/C\]. The time taken by the electron to fall this distance is:

A) \[1.3\times {{10}^{2}}s\]                     

B) \[2.1\times {{10}^{-12}}s\]

C) \[1.6\times {{10}^{-10}}s\]                  

D) \[2.9\times {{10}^{-9}}\,s\]

Correct Answer: D

Solution :

(d) \[2.9\times {{10}^{-9}}s\] Here, the direction of field upward. So, the negatively charged electron experiences a downward force. \[\therefore \]The acceleration of electron is \[{{a}_{e}}=\frac{eE}{{{m}_{e}}}\] The time required by the electron to fall through a distance h is \[{{t}_{e}}=\sqrt{\frac{2h}{{{a}_{e}}}}=\sqrt{\frac{2h{{m}_{e}}}{eE}}\]                    [using eq. (1)] \[={{\left[ \frac{2\times 1.5\times {{10}^{-2}}\times 9.11\times {{10}^{-31}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{4}}} \right]}^{{\scriptstyle{}^{1}/{}_{2}}}}=2.9\times {{10}^{-9}}5\]