(A) According to Bragg's equation: $n \lambda = 2 \ d \sin \theta$ Given: $n = 1$, $d = 280 \ pm = 280 \times 10^{-12} \ m$, $\sin \theta = 0.09$. Sub

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

(A) According to Bragg's equation: $n \lambda = 2 \ d \sin \theta$ Given: $n = 1$, $d = 280 \ pm = 280 \times 10^{-12} \ m$, $\sin \theta = 0.09$. Substituting the values: $\lambda = \frac{2 \times 280 \times 10^{-12} \ m \times 0.09}{1}$ $\lambda = 50.4 \times 10^{-12} \ m$ $\lambda = 0.504 \times 10^{-10} \ m = 0.504 \ \mathring{A}$

Class 12 Chemistry Solid State Mathematical analysis of cubic system and Bragg’s equation

Medium

In the $x$-ray reflection $(n=1)$, the distance between two parallel planes of $NaCl$ is $280 \ pm$ and the diffraction angle is $5.2^{\circ}$. What is the wavelength of its light radiation? (Given: $\sin 5.2^{\circ} = 0.09$)

AP EAMCET 2023 All AP EAMCET

A
$0.504 \ \mathring{A}$
B
$5.04 \ \mathring{A}$
C
$50.4 \ \mathring{A}$
D
$504 \ \mathring{A}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Solid State

Subtopic

Mathematical analysis of cubic system and Bragg’s equation

Previous Year

AP EAMCET 2023 Paper All AP EAMCET years →

$A$ metal has a $fcc$ lattice. The edge length of the unit cell is $404 \, pm$. The density of the metal is $2.72 \, g \, cm^{-3}$. The molar mass of the metal is :- ............ $g \, mol^{-1}$ ( $N_A$ Avogadro constant $= 6.02 \times 10^{23} \, mol^{-1}$ )

Medium

View Solution

The edge length of an $FCC$ unit cell is given by:

Easy

View Solution

For a crystal, the angle of diffraction $(2 \theta)$ is $90^{\circ}$ and the second order line has a $d$ value of $2.28 \ \text{Å}$. The wavelength (in $\text{Å}$) of $X$-rays used for Bragg's diffraction is

DifficultAP EAMCET 2008

View Solution

An element crystallizes in an $f.c.c.$ structure with a unit cell edge length of $200 \, pm$. If $200 \, g$ of this element contains $24 \times 10^{23}$ atoms,calculate its density in $\text{g cm}^{-3}$.

Difficult

View Solution

Ammonium chloride crystallizes in a body-centered cubic lattice with an edge length of the unit cell of $390 \ pm$. If the size of the chloride ion is $180 \ pm$,the size of the ammonium ion would be ........... $pm$.

DifficultAIEEE 2012

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

In the $x$-ray reflection $(n=1)$, the distance between two parallel planes of $NaCl$ is $280 \ pm$ and the diffraction angle is $5.2^{\circ}$. What is the wavelength of its light radiation? (Given: $\sin 5.2^{\circ} = 0.09$)

Similar Questions

$A$ metal has a $fcc$ lattice. The edge length of the unit cell is $404 \, pm$. The density of the metal is $2.72 \, g \, cm^{-3}$. The molar mass of the metal is :- ............ $g \, mol^{-1}$ ( $N_A$ Avogadro constant $= 6.02 \times 10^{23} \, mol^{-1}$ )

The edge length of an $FCC$ unit cell is given by:

For a crystal, the angle of diffraction $(2 \theta)$ is $90^{\circ}$ and the second order line has a $d$ value of $2.28 \ \text{Å}$. The wavelength (in $\text{Å}$) of $X$-rays used for Bragg's diffraction is

An element crystallizes in an $f.c.c.$ structure with a unit cell edge length of $200 \, pm$. If $200 \, g$ of this element contains $24 \times 10^{23}$ atoms,calculate its density in $\text{g cm}^{-3}$.

Ammonium chloride crystallizes in a body-centered cubic lattice with an edge length of the unit cell of $390 \ pm$. If the size of the chloride ion is $180 \ pm$,the size of the ammonium ion would be ........... $pm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module