Question

According to Moseley's law the square root of frequency of $K_{\alpha }$ characteristic X-ray is given by $\sqrt{v}=a(Z-b)$, What are the values of a & b?

• A. & 1
• B. & 1
• C. & 2
• D. & 2

Answer 1

The correct option is A & 1

$K_{\alpha }$ characteristic X-rays are emitted from an electronic transition from L-shell to K-shell.
Now let's look at it from the Bohr's model picture.
L-shell is obviously n=2, and K-shell is n=1. Hence we are getting an x-ray photon for electronic transition from n=2 to n=1.
We will follow the same algorithm to find the energy of a photon which we used to do while solving problems in Bohr's model of atom.
Now, we know that the problem in applying the Rydberg's formula here is that, Bohr's model was only used for a single electron system while here in our atom there are multiple electrons, hence we need to modify the Rydberg's formula. How do we do that?
Think when the electron is jumping from n=2 to n=1, between this electron and nucleus there is one more electron (The electron already present in n=1) now, this electron would screen the net effect of nucleus on the transitioning electron. How much? By a value equal to 1.
Which means in absence of that extra electron ,the transitioning electron would have felt the effect of Z protons but now will only feel the effect of Z-1 protons.
Hence we can also say that the value of effective atomic number is Z-1.
So, Energy of an atom with electron in nth orbit will be
$\frac{1}{\lambda }=R(Z-1{)}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
Now,
$n_1=1,n_2=2$
$\frac{1}{\lambda }=R(Z-1{)}^2$ $(1-\frac{1}{4})$
Hence,
$\frac{1}{\lambda }=R(Z-1{)}^{2}X\frac{3}{4}$
Or,
$\frac{1}{\lambda }=\frac{3R}{4}(Z-1{)}^2$
We need in terms of frequency hence;
$\frac{v}{c}=\frac{3R}{4}(Z-1{)}^2$
Or,
$v=\frac{3Rc}{4}(Z-1{)}^2$
Or,
$\sqrt{v}=\sqrt{\frac{3RC}{4}}(Z-1)$
Comparing with Moseley's law;
$a=\sqrt{\frac{3RC}{4}}$ and b=1.