If the resolution of a spectrograph is Delta lambda = 10^-12 m, would it be able...

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?

Solutions

Expert Solution

Using a formula, we have

E = - (13.6 eV) (Z - 1)² [(1 / n_upper²) - (1 / n_lower²)]

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)

genius_generous answered 4 months ago

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