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Determine the de Broglie wavelength of an electron that undergoes acceleration by a potential difference of 200 V. (Given:h= 6.626\times 10^{-3}Js,m_{e}= 9.1 \times 10^{-31}kg, e= 1.6\times 10^{-19}C )

Option: 1

1.23 \AA

Option: 2

0.616 \, \AA

Option: 3

2.46\, \AA

Option: 4

0.308\, \AA

Answers (1)


Using the equation for the de Broglie wavelength of an electron, we have:

\lambda_{e}= \frac{h}{\sqrt{2m_{e}\left ( eV \right )}}

Substituting the given values, we get:
\lambda_{e}= \frac{6.626\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 200}}

Solving this expression, we get:

\lambda_{e} \approx0.308\, \AA

The resulting expression is then evaluated to obtain the de Broglie wavelength of the electron. The correct answer is an option (4).  

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