Question

A diatomic molecule is made of two masses m₁ and m₂ which are separated by a distance r. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by (n is an integer)

$fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis squared straight n squared straight h squared over denominator 2 straight m subscript 1 superscript 2 straight m subscript 2 superscript 2 straight r squared end fraction$
$fraction numerator straight n squared straight h squared over denominator 2 left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight r squared end fraction$
$fraction numerator 2 straight n squared straight h squared over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight r squared end fraction$
$fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight n squared straight h squared over denominator 2 straight m subscript 1 straight m subscript 2 straight r squared end fraction$

Solution

fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight n squared straight h squared over denominator 2 straight m subscript 1 straight m subscript 2 straight r squared end fraction

Rotational kinetic energy of the two body system rotating about their centre of mass is
$RKE space equals space 1 half μω squared straight r squared where comma space straight mu space equals space fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction equals space reduced space mass and space angular space momentum comma space straight L space equals space μωr squared space equals space fraction numerator nh over denominator 2 straight pi end fraction straight omega space equals fraction numerator nh over denominator 2 πμr squared end fraction therefore space RKE space equals space 1 half μω squared straight r squared space equals space 1 half space straight mu. open parentheses fraction numerator nh over denominator 2 πμr squared end fraction close parentheses squared straight r squared space equals fraction numerator straight n squared straight h squared over denominator 8 straight pi squared μr squared end fraction space equals space fraction numerator straight n squared straight ħ squared over denominator 2 μr squared end fraction space open parentheses where comma straight ħ space equals space fraction numerator straight h over denominator 2 straight pi end fraction close parentheses fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight n squared straight ħ squared over denominator 2 straight m subscript 1 straight m subscript 2 straight r squared end fraction$

For Daily Free Study Material Join wiredfaculty

Units And Measurement

Some More Questions From Units and Measurement Chapter

Can we use different units to measure same physical quantity?

To measure a physical quantity X, the unit U is repeatedly used for n times. What is the magnitude of physical quantity?

Does the numerical factor in the quantitative measurement depend on the system of unit used?

Is one unit sufficient to measure a physical quantity?

Can a physical quantity have two or more units? Give example.

What are fundamental quantities?

How many fundamental quantities are there in mechanics?

How many fundamental quantities are there in Physics?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction