The K_{alpha} wavelength emitted by an atom o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to Moseley's law,the frequency of $K_{\alpha}$ $X$-rays is given by $
u = c/\lambda = a(Z - b)^2$,where $b = 1$ for $K_{\alpha}$ lines.

Vedclass · Accepted Answer

$Z = 6$

Vedclass · Answer

(A) According to Moseley's law,the frequency of $K_{\alpha}$ $X$-rays is given by $
u = c/\lambda = a(Z - b)^2$,where $b = 1$ for $K_{\alpha}$ lines.Thus,$\frac{1}{\lambda} \propto (Z - 1)^2$,or $\frac{1}{\sqrt{\lambda}} \propto (Z - 1)$.For the first atom,$Z_1 = 11$ and $\lambda_1 = \lambda$. So,$\frac{1}{\sqrt{\lambda}} = k(11 - 1) = 10k$.For the second atom,$Z_2 = Z$ and $\lambda_2 = 4\lambda$. So,$\frac{1}{\sqrt{4\lambda}} = k(Z - 1) = \frac{1}{2\sqrt{\lambda}} = k(Z - 1)$.Dividing the two equations: $\frac{1/\sqrt{\lambda}}{1/(2\sqrt{\lambda})} = \frac{10k}{k(Z - 1)}$.$2 = \frac{10}{Z - 1}$.$Z - 1 = 5$,which gives $Z = 6$.

The $K_{\alpha}$ wavelength emitted by an atom of atomic number $Z = 11$ is $\lambda$. Find the atomic number for an atom that emits $K_{\alpha}$ radiation with wavelength $4\lambda$.

Similar Questions

If $f_1, f_2$,and $f_3$ are the frequencies of $K_\alpha, K_\beta$,and $L_\alpha$ $X$-rays of a given metal,respectively,then:

$X$-rays cannot be deflected by means of an ordinary grating due to

If $X$-rays are produced by bombarding electrons with an energy of $40 \ keV$ on a target,what will be the minimum wavelength (in $\mathring{A}$) of the $X$-rays produced?

The minimum wavelength of $X$-rays produced in a Coolidge tube operating at $40 \ kV$ is ..... $\mathring A$.

What $kV$ potential is to be applied on an $X$-ray tube so that the minimum wavelength of emitted $X$-rays may be $1 \text{ Å}$? $(h = 6.625 \times 10^{-34} \text{ J-s})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module