By applying Moseley's law with b = 1 , calculate the frequency of the K_ X-ray of La ( Z = 57 ) given that the frequency of the K_ X-ray of Cu ( Z = 29 ) is 1.88 10¹⁸ Hz .

Step-by-step solution

According to Moseley's law, the frequency of K_ X-ray is given by: = a(Z - b) Given b = 1 , we have (Z - 1)^2 . For Cu ( Z = 29 ): _ Cu (29 - 1)^2 = 28^2 For La ( Z = 57 ): _ La (57 - 1)^2 = 56^2 Taking the ratio of the frequencies: _ La _ Cu = ( 56 28 )^2 = 2^2 = 4 _ La = 4 _ Cu Substituting the given value _ Cu = 1.88 10¹⁸ Hz : _ La = 4 1.88 10¹⁸ Hz = 7.52 10¹⁸ Hz