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Question
A proton and an α -particle are accelerated through the same potential difference. Which one of the two has greater de-Broglie wavelength Justify your answer.
Solution
The de Broglie wavelength is related to the accelerating potential as
`lambda=h/sqrt(2meV)`
`:.lambda_p=h/sqrt(2m_pe_pV)`
And `lambda_alpha=h/sqrt(2m_alphae_alphaV)`
`:.lambda_p/lambda_alpha=h/sqrt(2m_pe_pV)xxsqrt(2m_alphae_alphaV)/h=sqrt(m_alphae_alpha)/sqrt(m_pe_p)`
Now, mass and charge of α particle is greater than that of proton.
∴ λp > λα for the same potential difference.
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