Answer:
the equilibrium bond length of H137I is [tex]4.75*10^{-12}m[/tex]
Explanation:
The quantum state has a rotational energy which can be expressed mathematically
[tex]E_J =BJ(J+1)[/tex]
Here,
B is the rotational constant given as
[tex]B = \frac{h^2}{8r^2I} \ Joules[/tex]
We are given from the question that [tex]E_{J=5} = 8.952*10^{-21}J[/tex]
Substituting we have
[tex]8.952*10^{-21}kgm^2s^{-2} = B * 5(5+1)[/tex]
[tex]8.952*10^{-21}kgm^2s^{-2} =\frac{h^2}{8r^2I} *30[/tex]
Where,
h is plank constant
r is radius
I is moment of inertia
[tex]8.952*10^{-21}kgm^2s^{-2} = \frac{6.626*10^{-34}kgm^2s^{-2} }{8(3.142)^2*I} *30[/tex]
Making [tex]I[/tex] the subject
[tex]I = \frac{(6.626*10^{-34})^2 kgm^2s^{-2}}{8*(3.142)^2 * 8.952*10^{-21} kgm^2s^{-2}}[/tex]
[tex]= 3.701*10^{-47}kgm^2[/tex]
The moment of inertia = Reduced mass × [tex]\lambda^2[/tex]
[tex]\lambda \ is \ bond \ length[/tex]
[tex]Formula \ for\ reduced \ mass =\frac{m_1m_2}{\frac{1}{m_1} +\frac{1}{m_2} }[/tex]
[tex]mass =\frac{molar mass}{Avogadro's number}[/tex]
Therefore
[tex]m_1 = \frac{1}{6.022*10^{23}}[/tex]
[tex]m_2 = \frac{137}{6.022*10^{23}}[/tex]
[tex]I = \frac{\frac{1}{6.022*10^{23}} *\frac{137}{6.022*10^{23}} }{\frac{1+137}{6.022*10{23}} } * \lambda^2[/tex]
Making [tex]\lambda[/tex] the subject of the formula
[tex]\lambda^2 = \frac{3.701*10^{-47}}{1.64*10^{-24}}m^2[/tex]
[tex]\lambda = 4.75 *10^{-12} m[/tex]
