# Proton and an α -particle Are Accelerated Through the Same Potential Difference Which One of the Two Has Greater De-broglie Wavelength - Physics

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.

Concept: de-Broglie Relation
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