At what angles for the first order diffraction, spacing between two planes, respectively, are $\lambda $ and $\dfrac{\lambda }{2}$ ?
(A) ${0^ \circ },{90^ \circ }$
(B) ${90^ \circ },{0^ \circ }$
(C) ${30^ \circ },{90^ \circ }$
(D) ${90^ \circ },{30^ \circ }$
Answer
299.4k+ views
Hint: The equation that relates the interplanar distance and the angle of diffraction is the Bragg’s equation and it can be given as
\[2d\sin \theta = n\lambda \]
Complete step by step solution:
Bragg’s law gives the angles for the coherent and incoherent scattering of light from a crystal lattice. We know that in crystalline solid, the light waves are scattered from the lattice planes which are separated by the interplanar distance d.
- Scientist Bragg gave the relation between the path differences between the two waves undergo interference and diffraction angle. The Bragg’s equation is given as
\[2d\sin \theta = n\lambda \]
Where d is interplanar distance and n is a positive integer. $\lambda $ is the wavelength of the incident wave.
- We are provided with the question that the diffraction is of first order. So, the value of n is 1.
- Now, in one case, we are given that the interplanar distance is $\lambda $. So, in that case, the Bragg equation will be
\[2d\sin \theta = n\lambda \]
Putting the available values, we will get
\[2\lambda \sin \theta = (1)\lambda \]
So,
\[\sin \theta = \dfrac{\lambda }{{2\lambda }} = \dfrac{1}{2}\]
So, we can say that $\sin {30^ \circ } = \dfrac{1}{2}$ .
So, $\theta = {30^ \circ }$
In the second case, we are given that the interplanar distance is $\dfrac{\lambda }{2}$. So, putting this in the Bragg’s equation will give
\[2\left( {\dfrac{\lambda }{2}} \right)\sin \theta = (1)\lambda \]
So, we can write that
\[\sin \theta = \dfrac{{2\lambda }}{{2\lambda }} = 1\]
Thus, $\sin {90^ \circ } = 1$
So, we got that $\theta = {90^ \circ }$
Thus, we obtained that the angles will be ${30^ \circ }{\text{ and 9}}{{\text{0}}^ \circ }$ respectively.
Therefore, the correct answer is (C).
Note: Note that in the equation, n is the order of diffraction and it is always an integer value. With Bragg's law, we can find the lattice spacing for different cubic lattice systems which also includes the use of Miller indices.
\[2d\sin \theta = n\lambda \]
Complete step by step solution:
Bragg’s law gives the angles for the coherent and incoherent scattering of light from a crystal lattice. We know that in crystalline solid, the light waves are scattered from the lattice planes which are separated by the interplanar distance d.
- Scientist Bragg gave the relation between the path differences between the two waves undergo interference and diffraction angle. The Bragg’s equation is given as
\[2d\sin \theta = n\lambda \]
Where d is interplanar distance and n is a positive integer. $\lambda $ is the wavelength of the incident wave.
- We are provided with the question that the diffraction is of first order. So, the value of n is 1.
- Now, in one case, we are given that the interplanar distance is $\lambda $. So, in that case, the Bragg equation will be
\[2d\sin \theta = n\lambda \]
Putting the available values, we will get
\[2\lambda \sin \theta = (1)\lambda \]
So,
\[\sin \theta = \dfrac{\lambda }{{2\lambda }} = \dfrac{1}{2}\]
So, we can say that $\sin {30^ \circ } = \dfrac{1}{2}$ .
So, $\theta = {30^ \circ }$
In the second case, we are given that the interplanar distance is $\dfrac{\lambda }{2}$. So, putting this in the Bragg’s equation will give
\[2\left( {\dfrac{\lambda }{2}} \right)\sin \theta = (1)\lambda \]
So, we can write that
\[\sin \theta = \dfrac{{2\lambda }}{{2\lambda }} = 1\]
Thus, $\sin {90^ \circ } = 1$
So, we got that $\theta = {90^ \circ }$
Thus, we obtained that the angles will be ${30^ \circ }{\text{ and 9}}{{\text{0}}^ \circ }$ respectively.
Therefore, the correct answer is (C).
Note: Note that in the equation, n is the order of diffraction and it is always an integer value. With Bragg's law, we can find the lattice spacing for different cubic lattice systems which also includes the use of Miller indices.
Recently Updated Pages
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main

What is isoelectric point class 12 chemistry JEE_Main

Chlorobenzene is extremely less reactive towards a class 12 chemistry JEE_Main

In order to convert Aniline into chlorobenzene the class 12 chemistry JEE_Main

Give one chemical test to distinguish between the following class 12 chemistry JEE_Main

The coordination number of an atom in a fcc lattice class 12 chemistry JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
NCERT Solutions For Class 12 Chemistry Chapter 2 Electrochemistry - 2026-27 Free PDF Download (Sign-in Required)

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2026-27 Free PDF Download (Sign-in Required)

NCERT Solutions For Class 12 Chemistry Chapter 3 Chemical Kinetics - 2026-27 Free PDF Download (Login Required)

CBSE Notes Class 12 Chemistry Chapter 1 - Solutions - 2026-27 PDF Download (Login Required)

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes - 2026-27 Free PDF Download (Sign-In Required)

NCERT Solutions For Class 12 Chemistry Chapter 7 Alcohol Phenol And Ether - 2026-27 Free PDF Download (Login Required)

