The de Broglie wavelength associated with an electron accelerated through a potential difference V is λ_e and the de Broglie wavelength associated with a proton accelerated through the same potential difference is λ_p .

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Correct answer: A. (m_p)/(m_e)

The de Broglie wavelength of a particle accelerated through a potential difference

V

is given by

\lambda = \dfrac{h}{\sqrt{2mqV}}

.

For an electron, the charge is

e

and mass is

m_e

. Its de Broglie wavelength is

\lambda_e = \dfrac{h}{\sqrt{2m_e eV}}

.

For a proton, the charge is

e

and mass is

m_p

. Its de Broglie wavelength is

\lambda_p = \dfrac{h}{\sqrt{2m_p eV}}

.

Taking the ratio of the two wavelengths, we get

\dfrac{\lambda_e}{\lambda_p} = \dfrac{\dfrac{h}{\sqrt{2m_e eV}}}{\dfrac{h}{\sqrt{2m_p eV}}} = \sqrt{\dfrac{m_p}{m_e}}

.

Answer:

\sqrt{\dfrac{m_p}{m_e}}

The de Broglie wavelength associated with an electron accelerated...

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