Question

Use Moseley’s law with $b=1$, if the frequency of the $K_{\alpha }\text{X-ray}$ of $Cu(Z=29)$ is known to be $1.88\times {10}^{18}\text{Hz}$, then the frequency of the $K_{\alpha }\text{X-ray}$ of $La(Z=57)$ is $7.52\times {10}^{1x}\text{Hz}$. The value of $x$ is -

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

According to Moseley's law

$\sqrt{\nu }=a(Z-b)$

For $La(Z=57)$,

$\Rightarrow \sqrt{\nu }=a(57-1)=a\times 56...\left(1\right)$

For $Cu(Z=29)$,

$\sqrt{1.88\times {10}^{18}}=a(29-1)=28a...\left(2\right)$

Dividing (1) by (2),

$\sqrt{\frac{\nu }{1.88\times {10}^{18}}}=\frac{a\times 56}{a\times 28}=2$

$\Rightarrow \nu =1.88\times {10}^{18}\times (2{)}^2$

$=4\times 1.88\times {10}^{18}=7.52\times {10}^{18}\text{Hz}$

Given, $\nu =7.52\times {10}^{1x}\text{Hz}$

$\therefore x=8$