The energy difference of two stable orbits of | vedclass

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(A) Given: $\Delta E = 214.68 \ kJ \ mol^{-1} = 214.68 	imes 10^3 \ J \ mol^{-1}$.To find the energy per electron,divide by Avogadro's number $(N_A =

Vedclass · Accepted Answer

$5.39 	imes 10^{14} \ Hz$

Vedclass · Answer

(A) Given: $\Delta E = 214.68 \ kJ \ mol^{-1} = 214.68 	imes 10^3 \ J \ mol^{-1}$.To find the energy per electron,divide by Avogadro's number $(N_A = 6.022 	imes 10^{23} \ mol^{-1})$:$\Delta E_{	ext{per electron}} = \frac{214.68 	imes 10^3}{6.022 	imes 10^{23}} \approx 3.565 	imes 10^{-19} \ J$.Using the relation $\Delta E = hv$,where $h = 6.626 	imes 10^{-34} \ J \ s$:$v = \frac{\Delta E}{h} = \frac{3.565 	imes 10^{-19}}{6.626 	imes 10^{-34}} \approx 5.38 	imes 10^{14} \ Hz$.

The energy difference of two stable orbits of Hydrogen is $214.68 \ kJ \ mol^{-1}$. If the transition of an electron takes place,find the emitted frequency. (Note: $v = \Delta E/h$ and $h = 6.626 \times 10^{-34} \ J \ s$)

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