The second order Bragg's diffraction of X-ray | vedclass

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

Question

(D) The Bragg's equation is given by $2d \sin 	heta = n\lambda$.Given: Order of diffraction $n = 2$,wavelength $\lambda = 1 \ \mathring{A}$,and angle

Vedclass · Accepted Answer

$1.15$

Vedclass · Answer

(D) The Bragg's equation is given by $2d \sin 	heta = n\lambda$.Given: Order of diffraction $n = 2$,wavelength $\lambda = 1 \ \mathring{A}$,and angle $	heta = 60^\circ$.Substituting the values into the equation:$2 	imes d 	imes \sin \ 60^\circ = 2 	imes 1 \ \mathring{A}$Since $\sin \ 60^\circ = \frac{\sqrt{3}}{2} \approx 0.8660$,we have:$2 	imes d 	imes 0.8660 = 2$$d = \frac{2}{2 	imes 0.8660} = \frac{1}{0.8660} \approx 1.1547 \ \mathring{A}$.Rounding to two decimal places,$d = 1.15 \ \mathring{A}$.

The second order Bragg's diffraction of $X$-rays with $\lambda = 1 \ \mathring{A}$ from a set of parallel planes in a metal occurs at an angle of $60^\circ$. The distance between the scattering planes in the crystal is ................ $\mathring{A}$.

Similar Questions

If the edge length of the unit cell of an element crystallizing in a $bcc$ structure is $400 \, pm$, then the density of the crystal will be ............ $\text{g/cm}^3$. (Atomic weight $= 100 \, \text{g/mol}$)

Calculate the volume of $fcc$ unit cell if the radius of a particle in it is $106.05 \ pm$.

Lithium forms a body-centred cubic $(BCC)$ structure. The length of the side of its unit cell is $351 \ pm$. The atomic radius of lithium will be: ............. $pm$

Calculate the radius of a metal atom if it forms a $bcc$ unit cell having an edge length of $530 \ pm$. (in $pm$)

Copper crystallises in $fcc$ with a unit cell edge length of $361 \ pm$. What is the radius of the copper atom in $pm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module