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← JEE Main 2026 · 4 Apr · Shift 2
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JEE Main 2026 · 4 Apr · Shift 2, Physics question 44

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 and worked solution

Official answer: A

de Broglie wavelength. λ=h2mqV\lambda=\dfrac h{\sqrt{2mqV}}; same q and V, so λeλp=mpme\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}.

(A)