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.

Answer 1

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.