1.

Calculate the ratio of the accelerating potential required to accelerate (i) a proton and (ii) and `alpha` particle to have the same de-Broglie wavelength associated with them.

Answer» If the charged particle of mass m, charge q is accelerated under a pot. diff. V, the velocity acquired by particle is v. Then
`qV=1/2mv^2=(m^2v^2)/(2m)=(p^2)/(2m)`
Where p is momentum of particle
of `p=sqrt(2mqV)`
De-Broglie wavelength, `lambda=h/p=h/(sqrt(2mqV))`
Now, `lambda_p=h/(sqrt(2m_pq_pV_p)) and lambda_(alpha)=h/(sqrt(2m_(alpha)q_(alpha)V_(alpha)))`
of `(V_p)/(V_(alpha))=(m_(alpha)q_(alpha))/(m_pq_p)=(4mxx2e)/(mxxe)=8`


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