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  • An X-ray tube operates at . Then the maximum speed of the electrons striking the target is : (

An X-ray tube operates at 5 \mathrm{KV}. Then the maximum speed of the electrons striking the target is : (Given: Mass of electron \mathrm{m}=9.1 \times 10^{-31} \mathrm{Kg}, charge on electron, e=1.602 \times 10^{-19} \, C.)

Option: 1

4.2 \times 10^7 \mathrm{~m} / \mathrm{s}.


Option: 2

3.2 \times 10^7 \mathrm{~m} / \mathrm{s}.


Option: 3

2.1 \times 10^7 \mathrm{~m} / \mathrm{s}.


Option: 4

None of these


Answers (1)

best_answer

Applied voltage, \mathrm{V}=5 \mathrm{KV}=5 \times 10^3 \, \mathrm{Volts}.

Let \mathrm{v} \ \mathrm{m} / \mathrm{s} represents the speed the electrons at which they strike the target.
Kinetic energy of the electron =(1 / 2) \mathrm{mv}^2

Now, kinetic energy gained by the electron is falling through a potential difference of V volts is given by \Delta \mathrm{K} . \mathrm{E} .=\mathrm{eV}

Hence,\mathrm{eV}=(1 / 2) \mathrm{mv}^2 or  \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}=\sqrt{\frac{2 \times 1.602 \times 10^{-19} \times 5 \times 10^3}{9.1 \times 10^{-31}}}

=4.2 \times 10^7 \mathrm{~m} / \mathrm{s}.

Posted by

Ritika Kankaria

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