1.

lamda_pand lamda_(alpha) be the wavelengths of proton and alpha-particle of equal kinetic energy, then

Answer»

`lamda_p = (lamda_(alpha))/4`
`lamda_p = (lamda_(alpha))/2`
`lamda_p = lamda_(alpha)`
`lamda_p = 2lamda_(alpha)`

SOLUTION :de Broglie wavelength , `LAMDA = h/(sqrt(2mK))`
m = MASS of a particle
K = kinetic energy of a particle
For proton`lamda_p=h/(sqrt(2m_pK_p)) ""...(i)`
For `alpha` - particle , `lamda _(alpha) = (h)/(sqrt(2m_(alpha)K_(alpha))`
`lamda_(alpha) = h/(sqrt2(4m_p) K_p)"" ( :. K_p = K_(alpha)) `
`lamda_(alpha) = h / (2 sqrt(2m_p K_p))""...(ii)`
DIVIDING (i) by (ii) , we get
`lamda_p/lamda_(alpha)=(h/(sqrt(2m_pK_p)))/(h/(2sqrt(2m_pK_p)))`
`lamda_p/lamda_(alpha)=2, " or " lamda_(p) = 2lamda_(alpha)`


Discussion

No Comment Found

Related InterviewSolutions