The wavelength of the K_alpha line for an ele | vedclass

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

Question

(C) According to Moseley's law,the frequency $
u$ of the $K_\alpha$ line is given by $
u = cR(Z-b)^2(\frac{1}{1^2} - \frac{1}{2^2})$,which implies $

Vedclass · Accepted Answer

$4\lambda$

Vedclass · Answer

(C) According to Moseley's law,the frequency $
u$ of the $K_\alpha$ line is given by $
u = cR(Z-b)^2(\frac{1}{1^2} - \frac{1}{2^2})$,which implies $\frac{1}{\lambda} \propto (Z-b)^2$.For the first element with $Z_1 = 29$ and $b = 1$,we have $\frac{1}{\lambda_1} = k(29-1)^2 = k(28)^2$,where $\lambda_1 = \lambda$.For the second element with $Z_2 = 15$ and $b = 1$,we have $\frac{1}{\lambda_2} = k(15-1)^2 = k(14)^2$.Taking the ratio of the two equations:$\frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - b)^2}{(Z_2 - b)^2} = \frac{(29-1)^2}{(15-1)^2} = \frac{28^2}{14^2} = (\frac{28}{14})^2 = 2^2 = 4$.Therefore,$\lambda_2 = 4\lambda_1 = 4\lambda$.

The wavelength of the $K_\alpha$ line for an element of atomic number $Z_1 = 29$ is $\lambda$. What is the wavelength of the $K_\alpha$ line for an element of atomic number $Z_2 = 15$? (Take Moseley's constant $b = 1$ for both elements.)

Similar Questions

$X$-rays are produced in an $X$-ray tube operating at a given accelerating voltage. The wavelength of the continuous $X$-rays has values from

The energy of the $K_\alpha$ $X$-ray line of an element is $8578$ units,and the energy of the $L$ level is $-2859$ units. The energy of the $K$ level is . . . . . . units.

If the wavelength of the $K_\alpha$ line of an element with atomic number $43$ is $\lambda$,then the wavelength of the $K_\alpha$ line of an element with atomic number $29$ is:

Characteristic $X$-rays are produced due to

The energy of a photon of characteristic $X-$ rays from a Coolidge tube comes from

Vedclass Test Series

Exam Paper Generator

Online Exam Module