Question

problem 20-7

x modifier in atomic 20- ctroscopy? The first excited state of Ca is reached by absorption each cur trati of 422.7-nm light. hat is the energy difference (0) between the ground and cited states? (Hint: See Section 18-1.) b) The degeneracies are g"/g0 3 for Ca. Find N*/No at 2500 K. (Hg By what percentage will the fraction in (b) be changed by a 15-K rise in temperature? (d) Find N*/No at 6 000 K. 20-7. The first excited state of Cu is reached by absortion20- of 327-nm radiation. (a) What is the energy difference (J) between the ground and K ser excited states? add (b) The ratio of degeneracies is g*/go 3 for Cu. Find N*INo ture at 2 400 K. (c) By what percentage will the fraction in (b) be changed by miv a 15-K rise in temperature? (d) What will the ratio N*/No be at 6 000 K? The Detection limit. (Refer to Section 5-2.) To estimate K nal 20- detection limit for arsenic in tap water by furnace atomic rption, eight tap water samples with a low concentration rAs were measured. Then eight tap water samples spiked abso Na con 0 ppb (ng/mL) As were measured.

Answer 1

A ) First we need to calculate the amount of energy per Photon

E=h$\nu$ Where h is the planks constant and $\nu$ isthe frequency of the radiation which is equal to $\frac{C}{\lambda }$ nm

E= $\frac{hc}{\lambda }$ Or E = $\frac{6.62 \times 10^{-34} J-s \times 3.00 \times 10^{8} m/s}{327 nm\times 10^{-9} }$ = 6.02 J/photon

10-19

Since the ground state contain 1 mole or avagadro number of partcle

the total energy is equal to $6.02\times 10^{-19} \times 6.02\10^{23}$    = 3.64$\times 10^{5}$ J/Photon

Or E = 364 kJ/mol

b)