एक द्विपरमाणुक अणु `m_(1)` तथा `m_(2)` के दो द्रव्यमानों से बना है जो कि दूरी पर है | यदि कोणीय संवेग क्वाण्टीकरण के बोर के नियम को लगाकर हम इसकी घूर्णन ऊर्जा की गणना करें, तब इसकी ऊर्जा दी जाएगी (n एक पूर्णांक है)

Options

A. `((m_(1)+m_(2))^(2)n^(2)h^(2))/(2m_(1)^(2)m_(2)^(2)r^(2))`
B. `(n^(2)h^(2))/(2(m_(1)+m_(2))r^(2))`
C. `(2n^(2)h^(2))/((m_(1)+m_(2))r^(2))`
D. `((m_(1)+m_(2))n^(2)h^(2))/(2m_(1)m_(2)r^(2))`

Answer

D

Explanation

`m_(1)r_(1)=m_(2)r_(2)`
`r_(1)+r_(2)=r`
`therefore r_(1)=(m_(2)r)/(m_(1)+m_(2)),r_(2)=(m_(1)r)/(m_(1)+m_(2))`
`therefore epsilon=(1)/(2)Iomega^(2)`
`=(1)/(2)(m_(1)r_(1)^(2)+m_(2)r_(2)^(2)).omega^(2)`
`mvr=(nh)/(2pil)`
`therefore epsilon=(1)/(2)I.(n^(2)h^(2))/(4pi^(2)I^(2))`
`=(n^(2)h^(2))/(8pi^(2))(1)/(m_(1)(m_(2)^(2)r^(2))/((m_(1)+m_(2))^(2))+m_(2)(m_(1)^(2)r^(2))/((m_(1)+m_(2))^(2)))`
`=(n^(2)h^(2))/(8pi^(2)r^(2))((m_(1)+m_(2))^(2))/(m_(1)m_(2)(m_(1)+m_(2)))=((m_(1)+m_(2))n^(2)h^(2))/(8pi^(2)r^(2)m_(1)m_(2))`
`=((m_(1)+m_(2))n^(2)h^(2))/(2m_(1)m_(2)r^(2))`