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Question
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.
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 mp and e respectively, and, mass and charge of a deutron are 2mp 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.
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