If the wavelength of the K_alpha line of an e | 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 = c/\lambda = R(Z-1)^2(1/1^2 - 1/2^2)$.Thus,$1/\lambda \pro

Vedclass · Accepted Answer

$(9/4) \lambda$

Vedclass · Answer

(C) According to Moseley's Law,the frequency $
u$ of the $K_\alpha$ line is given by $
u = c/\lambda = R(Z-1)^2(1/1^2 - 1/2^2)$.Thus,$1/\lambda \propto (Z-1)^2$.Let $\lambda_1$ be the wavelength for $Z_1 = 43$ and $\lambda_2$ be the wavelength for $Z_2 = 29$.Then,$\lambda_2 / \lambda_1 = (Z_1 - 1)^2 / (Z_2 - 1)^2$.Substituting the values: $\lambda_2 / \lambda = (43 - 1)^2 / (29 - 1)^2$.$\lambda_2 / \lambda = (42)^2 / (28)^2 = (42/28)^2$.Simplifying the fraction $42/28 = 3/2$.Therefore,$\lambda_2 / \lambda = (3/2)^2 = 9/4$.Hence,$\lambda_2 = (9/4) \lambda$.

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:

Similar Questions

The electron transition required for the emission of characteristic $K_\beta$ $X$-rays is:

The wavelength of $K_\alpha$ $X$-rays produced in an $X$-ray tube is $0.76 \ \mathring{A}$. The atomic number of the anode material is:

Assertion : Soft and hard $X-$ rays differ in frequency as well as velocity.
Reason : The penetrating power of hard $X-$ rays is more than the penetrating power of soft $X-$ rays.

An $X$-ray tube operates at $30 \ kV$. What is the minimum wavelength emitted in $\mathring{A}$? (Given: $h = 6.6 \times 10^{-34} \ J \cdot s$,$e = 1.6 \times 10^{-19} \ C$,$c = 3 \times 10^8 \ m/s$)

The continuous $X-$ray spectrum produced by an $X-$ray machine at constant voltage has

Vedclass Test Series

Exam Paper Generator

Online Exam Module