यदि रदरफोर्ड के प्रयोग में `90^(@)` के कोण पर प्रकीर्णित होने वाले `alpha-` कणों की संख्या 56 प्रति मिनट हो तो `60^(@) "तथा" 120^(@)` के कोणों पर प्रकीर्णित होने वाले कणों की संख्याएँ ज्ञात कीजिए ।

Explanation

दिया है : `theta_(1) = 90^(@) , N_(1) = 56` प्रति मिनट
`theta_(2) = 60^(@) , N_(2) = ?`
`theta_(3) = 120^(@), N_(3) = ?`
`because N prop (1)/(sin^(4)((theta)/(2)))`
`therefore (N_(2))/(N_(1)) = [sin (theta_(1)/(2))/(sin ((theta_(2))/(2)))]^(4) = [sin(90^(@)/(2))/(sin(60^(@)/(2)))]^(4)`
या `(N_(2))/(N_(1)) = `[(sin45^(0))/(sin30^(@))]^(4)`
या `(N_(2))/(N_(1)) = `[((1)/(sqrt(2)))/((1)/(2))]^(4) = [(sqrt(2)]^(4) = 4`
`therefore N_(2) = 4N_(1)`
`= 4xx56 = 224` प्रति मिनट
इसी प्रकार `(N_(3))/(N_(1)) = [ sin ((90)/(2))/(sin (120^(0)/(2)))]^4 = [(sin45^(@))/(sin60^(@))]`
`(N_(3))/(N_(1)) = [((1)/(sqrt(2)))/(sqrt(3)/(2))]^(4) = [[sqrt(2))/(sqrt(3))]^(4) = (4)/(9)`
`therefore N_(3) = N_(1)xx (4)/(9) = (56 xx 4)/(9) = (224)/(9) ~~ 25`
या `N_(3) = 25` प्रति मिनट