KVPY SA Exam 2013 Question Paper

Question : 70 of 80

Marks: +1, -0

An electron with an initial speed of

4.0 × 10^6 \ms^{-1}

is brought to rest by an electric field. The mass and charge of an electron are

9 ×10^{-31} \kg

and

9 ×10^{-19} \C

, respectively. Identify the correct statement.

The electron moves from a region of lower potential to higher potential through a potential difference of 11.4 µV.
The electron moves from a region of higher potential to lower potential through a potential difference of 11.4 µV.
The electron moves from a region of lower potential to higher potential through a potential difference of 45 V.
The electron moves from a region of higher potential to lower potential through a potential difference of 45 V.

Solution:

\qV=1/2\m\v^2

\V=1/2{\m\v^2}/\q

\V=1/2×{9×10^{-31}×(4×10^6)^2}/{1.6×10^{-19}}=45\V

45 V from higher to lower potential.

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