Question

The vibrational frequency for a diatomic molecule HF is 2.24 X 1014 Hz The mass of hydrogen atom and fluorine atom are 1.60 x10 -27 Kg and

3.15 X 10-26 Kg Find (a) force constant K for the interatomic force

(b) The energies in ground state and first excited state.​

Answer 1

the answer is already given

Answer 2

Answer:

The force constant K for the interatomic force is equal to 1342kgs⁻².              

Energy of ground state = 7.42×10²⁰J

Energy of first excited state = 22.26×10⁻²⁰J

Explanation:

Given that,

•  vibrational frequency of diatomic molecule=  2.24×10¹⁴Hz

         From Hooke's law,       $v=\frac{1}{2\pi } \sqrt\frac{k}{\mu}$

          mass of H-atom = 1.60 ×10⁻²⁷kg

           mass of fluorine atom = 3.15× 10⁻²⁶kg

        Reduced mass,  $\mu=\frac{m_{H}m_{F}}{m_{H}+m_{F} }$

                                 $\mu =\frac{(1.60*10^{-27})({3.15*10^{-26})}}{(3.15*10^{-26})+(1.60*10^{-27})} \\ \\ \mu= 0.152*10^{-26}$

        Force constant,     $k=4\pi ^{2} \nu^{2} \mu$

                                  $k=4(3.14)^2( 2.24*10^{14} )(0.152*10^{-26})kgs^{-2}\\ \\ k=1342kgs^{-2$

• Energy of diatomic molecule, $E=(n+\frac{1}{2} )h\nu$

       For ground state, n=0  then

                                $E= \frac{1}{2} h\nu=\frac{1}{2}(6.626*10^{-34}Js^{1}) (2.24*10^{14}s)\\ \\ E=7.74*10^{-20}J$

      For first excited state, n=1

                            $E= \frac{3}{2} h\nu=\frac{3}{2}(6.626*10^{-34}Js^{1}) (2.24*10^{14}s)\\ \\ E=22.26*10^{-20}J$