Find the ratio of de Broglie wavelength of a proton and as `alpha`-particle which have been accelerated through same potential difference.

Options

A. ` 2 sqrt3`
B. `3 sqrt2`
C. `2 sqrt2`
D. `3sqrt3`

Answer

C

Explanation

K.E. gained by a charge q after being accelerated through a potential difference V volt is given by `qV = 1/2 mv^2`
`v = sqrt((2qV)/m)and mv=sqrt(2mqV)`
de Broglie wavelength , `lamda = h/ (mv) = h / sqrt(2mqV)`
Now ,` lamda_p=h/(sqrt(2m_pq_pV_p)`
For `alpha ` - particle
`lamda_(alpha)=h/sqrt(2m_alphaq_alphaV_alpha):. lamda_p/lamda_(alpha)= sqrt((m_alphaq_alphaValpha)/(m_pq_pV_p))`
Putting `V_alpha = V_p` , we get
`lamda_p/lamda_(alpha)=sqrt((m_alphaq_alpha)/(m_pq_p))=sqrt((4xx2)/(1xx1))=sqrt8=2sqrt2`