Question

The $X-$ ray wavelength of $L_{\alpha }$ line of Platinum $(Z=78)$ is $1.30A^0$. The $X-$ ray wavelength of $L_{\alpha }$ line of Molybdenum $(Z=42)$ is (constant $b=7.4$)

• A. $5.41A^0$
• B. $4.20A^0$
• C. $2.70A^0$
• D. $1.35A^0$

Answer 1

The correct option is A $5.41A^0$

By Moseley's law

$\sqrt{v}=a(z-b)$

a, b are constants far one type of x-ray's & for all elements (independent of z)

It is derived from Borh's theory:-

$\frac{1}{h}=R{z}^2(\frac{1}{h_1^2}-\frac{1}{h_2^2})$

for multi electron system z will be $(z-b)$ due to "shielding effect".

so, $\frac{1}{\lambda }=R(z-b{)}^2(\frac{1}{h_1^2}-\frac{1}{h_2^2})$

for $L_{\alpha }$ $h_1=2$, $n_2=3$

$\frac{h_2}{h_1}=\frac{(z_1-b{)}^2}{(z_2-b{)}^2}$

$z_1=78$

$z_2=42$

$b=7.4$

Given in question

$h_2=\frac{78-7.4{)}^2}{(42-7.4{)}^2}\times 1.3\stackrel{o}{A}=5.41\stackrel{o}{A}$

Option 'A' is right.