Dual Nature of Matter

JEE Main level, single correct, medium

8 of 200

Question 8 of 200

The de Broglie wavelength associated with an electron accelerated through a potential difference V is λe\lambda_e and the de Broglie wavelength associated with a proton accelerated through the same potential difference is λp\lambda_p. If their corresponding masses are mem_e and mpm_p, respectively, then the ratio of their de Broglie wavelengths (λeλp)\left(\dfrac{\lambda_e}{\lambda_p}\right) is ______.

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Answer: (A) mpme\sqrt{\dfrac{m_p}{m_e}}

The de Broglie wavelength of a particle accelerated through a potential difference VV is given by λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}.

For an electron, the charge is ee and mass is mem_e. Its de Broglie wavelength is λe=h2meeV\lambda_e = \dfrac{h}{\sqrt{2m_e eV}}.

For a proton, the charge is ee and mass is mpm_p. Its de Broglie wavelength is λp=h2mpeV\lambda_p = \dfrac{h}{\sqrt{2m_p eV}}.

Taking the ratio of the two wavelengths, we get λeλp=h2meeVh2mpeV=mpme\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: mpme\sqrt{\dfrac{m_p}{m_e}}

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