Part 1 and given So the example is as follows: h was given (6.26*10^-34 J*s) Rh=
ID: 1016751 • Letter: P
Question
Part 1 and given So the example is as follows: h was given (6.26*10^-34 J*s) Rh= 2.18*10^-18 E= (2.18*10^-18 J) (1/n^2-1/3^2)
Part 2 Equation: E=hv Subsituting numerical values in for energy and Planck's constant we get 1.94* 10^-18 J=(6.626*10^-34 J*s)=v
Solving for frequency gives 2.93*10^15s^-1=v
My question: So the n's were given in the question the initial was 1 and the final n was 3, but I was actually wondering how to calculate in part two. I dont understand how they got from 1.94* 10^-18 J=(6.626*10^-34 J*s)=v to 2.93*10^15s^-1=v
Explanation / Answer
E= (2.18*10^-18 J) (1/n^2-1/3^2)
n= 1 for initial value
E= (2.18*10^-18 J) (1/1^2-1/3^2)
E= (2.18*10^-18 J) (1/1-1/9)
E= (2.18*10^-18 J) (1-0.11)
= 2.18*10-18 *0.89 = 1.94*10-18 joules
E = hv
v = E/h
= 1.94*10-18/6.626*10-34
= 2.93*1015 sec-1 >>>> answer