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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.

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#### 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.

∴ λ_{p} > λ_{α} for the same potential difference.

Concept: de-Broglie Relation

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