Is that 138 THz a frequency. Change to energy, E = h*frequency. Then
f/c = 1.097E7*(1/4^2 - 1/n^2)
Solve for n. c = 3E8. f = frequency.
An electron in a hydrogen atom relaxes to the n= 4 level, emitting light of 138 THz .
What is the value of for the level in which the electron originated?
Please show each step and the equation.
2 answers
-2.18*10^-18(1/initial n^2 - final n^2) = hc/lamda.
h= 6.626*10^-34 it is a constant
c= 3.^*10^8 or 2.99*10^8 also a constant
lamda= 138 = 138*18^12
final n= 4
plug all the numbers i and solve for initial n
h= 6.626*10^-34 it is a constant
c= 3.^*10^8 or 2.99*10^8 also a constant
lamda= 138 = 138*18^12
final n= 4
plug all the numbers i and solve for initial n