ANY HELP IS MUCH APPRECIATED, I ALWAYS RATE, THANKYOU!!! . A series of lines in

Question

ANY HELP IS MUCH APPRECIATED, I ALWAYS RATE, THANKYOU!!! . A series of lines in the atomic spectrum of hydrogen lies at656.46 nm, 486.27 nm, 434.17nm, and 410.29 nm. What is thewavelength in the next line in the series? What is the ionizationenergy of the atom when it is in the lower state of thetransitions? . ANY HELP IS MUCH APPRECIATED, I ALWAYS RATE, THANKYOU!!! . A series of lines in the atomic spectrum of hydrogen lies at656.46 nm, 486.27 nm, 434.17nm, and 410.29 nm. What is thewavelength in the next line in the series? What is the ionizationenergy of the atom when it is in the lower state of thetransitions? .

Explanation / Answer

v= 1/l = RH(1/n22 -1/n12)

Thereforeby graphing 1/l vs. 1/n12,the following dataare obtained and the use of linear regression to find the nextwavelength, l.
yint =RH/n22
By solving forn2 in the yint, we find n2 =2

v= RH(1/n22 -1/n12) for n1 = 7 andn2 = 2, we find v = 23181.4cm-1
therefore l = 397.12 nm

IonizationEnergy = RHhc/n22 = 27419.85.45e-19 J

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