An electron of mass 9 - 10-31 kg moves with a speed of 107 m-s- It acquires a K-E- of-

The correct option is A 281.25 eV K.E. = gt; 12 mv^2 = gt; 12× 9×10^ 31× (10^7)^2 = gt; 4.5×10^ 17J #160; #160; #160; # ...

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Standard X

Physics

Kinetic Energy

An electron o...

Question

An electron of mass $9 \times 10^{- 31}$ kg moves with a speed of $10^{7}$ m/s. It acquires a K.E. of:

562.50 e V

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1125 e V

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1250 e V

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281.25 e V

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Solution

The correct option is A $281.25 e V$
K.E. => $\frac{1}{2} m v^{2}$
=> $\frac{1}{2} \times 9 \times 10^{- 31} \times (10^{7})^{2}$
=> $4.5 \times 10^{- 17} J$ ( $∵ 1.6 \times 10^{- 19} e V = 1 J$ )
=> $\frac{4.5 \times 10^{- 17}}{1.6 \times 10^{- 19}} e V$
=> $2.8125 \times 10^{2} e V$
=> $281.25 e V$ .

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An electron of mass 9×10−31 kg moves with a speed of 107 m/s. It acquires a K.E. of:

The correct option is A 281.25 eVK.E. => 12mv2=> 12×9×10−31×(107)2=> 4.5×10−17J (∵1.6×10−19eV=1J)=> 4.5×10−171.6×10−19eV=> 2.8125×102eV=>281.25 eV.

An electron of mass $9 \times 10^{- 31}$ kg moves with a speed of $10^{7}$ m/s. It acquires a K.E. of: