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A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has the greater value of de-Broglie wavelength associated with it, and Give reasons to justify your answer.

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#### Solution

de-Broglie wavelength of a particle depends upon its mass and charge for same accelerating potential, such that

`lambdaprop1/sqrt((`

Mass and charge of a proton are m_{p} and e respectively, and, mass and charge of a deutron are 2m_{p} and e respectively. where, e is the charge of an electron

`lambda_p/lambda_D=sqrt((m_Dq_D)/(m_pq_p))=sqrt(((2m_p)(e))/((m_p)(e)))=sqrt2`

Thus, de-broglie wavelength associated with proton is`sqrt2` times of the de-broglie wavelength of deutron and hence it is more.

Concept: de-Broglie Relation

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