If the first emission in the Lyman series of | vedclass

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(C) The energy difference $\Delta E$ for one mole of photons is given by $\Delta E = \frac{N_A hc}{\lambda}$.Given $\lambda = 121.5 \ nm = 121.5 	ime

Vedclass · Accepted Answer

$982$

Vedclass · Answer

(C) The energy difference $\Delta E$ for one mole of photons is given by $\Delta E = \frac{N_A hc}{\lambda}$.Given $\lambda = 121.5 \ nm = 121.5 	imes 10^{-9} \ m$,$h = 6.626 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.$\Delta E = \frac{6.022 	imes 10^{23} 	imes 6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{121.5 	imes 10^{-9}} \ J/mol$.$\Delta E \approx \frac{11.97 	imes 10^1}{121.5 	imes 10^{-9}} 	imes 10^{-11} \approx 984,000 \ J/mol$.Rounding to the nearest provided option,$\Delta E \approx 982 \ kJ/mol$.

If the first emission in the Lyman series of a hydrogen atom occurs at $121.5 \ nm$,what will be the energy difference between the first and second orbits in $kJ/mol$?

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