In: Physics
If the resolution of a spectrograph is Delta lambda = 10^-12 m, would it be able to separate the K(alpha) lines of platinum and gold?
Using a formula, we have
E
= - (13.6 eV) (Z - 1)2 [(1 /
nupper2) - (1 /
nlower2)]
The K
energies of Pt and Au which are given below as -
For platinum ; E,Pt
= - (13.6 eV) (78 - 1)2 [1 / (2)2 - 1 /
(1)2)]
E,Pt
= 60475.8 eV
For gold ; E,Au
= - (13.6 eV) (79 - 1)2 [1 / (2)2 - 1 /
(1)2)]
E,Au
= 62056.8 eV
We know that,
= h c /
E
= [(6.63x 10-34 J.s) (3 x 108 m/s)] /
[(62056.8 eV) - (60475.8 eV)]
= [(1.989 x 10-25 J.m) / (1581 eV)
converting eV into J :
= [(1.989 x 10-25 J.m) / (2.53 x 10-16
J)]
= 7.86 x 10-10 m
(Since this is larger than 10-12 m, then I will be able to see them as separate)