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Test: Bragg’s Law - Chemistry MCQ


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10 Questions MCQ Test Physical Chemistry - Test: Bragg’s Law

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Test: Bragg’s Law - Question 1

In the X-ray diffraction of a set of crystal planes having d equal to 0.18 nm, the first-order reflection is found to be at an angle of 22°. The wavelength of X-rays is: (sin 22° = 0.208)

Detailed Solution for Test: Bragg’s Law - Question 1

Concept:
Bragg’s Law states that when an X-ray beam are allowed to scatter through a crystalline surface, its angle of incidence θ will reflect with some angle of scattering θ. And when the path difference d is equal to a whole number n, of wavelength.
It can be represented as
2dsinθ  =nλ
Calculation:
Given:
d = 0.18 nm, n = 1, θ = 22°
⇒ 2 × 0.18 × sin 22° = 1 ⋅ λ
⇒ λ = 0.07488 nm

Test: Bragg’s Law - Question 2

If an X-ray tube operates at the voltage of 10 kV, find the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced. The specific charge of an electron is 1.8 × 1011 C / kg.

Detailed Solution for Test: Bragg’s Law - Question 2

CONCEPT:
The wavelength of any charged particle due to its motion is called the de-Broglie wavelength.
When a charged particle is accelerated in a potential difference the energy gained by the particle is given by:
Energy (E) = e × V
Where V is the potential difference and q is a charge.
The de-Broglie wavelength of charge particle (λd) is given by:

Where E is energy, h is Planck constant, m is the mass of the charged particle
EXPLANATION:
De-Broglie wavelength when a charge q is accelerated by a potential difference of V volts is

For cut off wavelength of X-rays, we have


From Eqs, i) and ii), we get

For electron q/m = 1.8 x 1011C/kg (given).
Substituting the values the desired ratio is

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Test: Bragg’s Law - Question 3

In the X-ray diffraction of a set of crystal planes having d equal to 0.18 nm, first order reflection is found to be at an angle of 22°. The wavelength of X-rays is: (sin 22° = 0.374, cos 22° = 0.927)

Detailed Solution for Test: Bragg’s Law - Question 3

Concept:
Bragg’s Law states that when an X-ray beam are allowed to scatter through a crystalline surface, its angle of incidence θ will reflect with some angle of scattering θ. And when the path difference d is equal to a whole number n, of wavelength.
It can be represented as
2dsinθ  =nλ
Calculation:
Given:
d = 0.18 nm, n = 1, θ = 22°
From Bragg’s law:
2d sin θ = nλ
2 × 0.18 × sin 22° = 1 ⋅ λ
λ = 0.13464 nm

Test: Bragg’s Law - Question 4

X-rays can be deflected by:

Detailed Solution for Test: Bragg’s Law - Question 4

X-rays aren't deflected by electrons and magnetic  fields because x-rays do not carry and charge. They  are electro-magnetic radiations and therefore  cannot be deflected by electronic or any magnetic  fields. 
EXTRA POINTS:
1. X-rays: An electromagnetic wave with wavelengths from 0.1 Å to 100 Å called X-rays.
2. X-rays were discovered by WC Roentgen in 1895 and he was later awarded with Nobel Prize in 1901. So option 4 is correct.
3. He named that new unknown type of radiation as X-rays.
4. X-rays are the second most energetic radiations in the Electromagnetic Spectrum following Gamma rays.
5. The wavelength of these rays ranges from 10-11 to 10-8 m and the frequency is around 3 x 1016 - 3 x 1019.
6. Therefore these are very shortwave and high energy EM radiations (travelling with the speed of light).
7. X-rays are widely used in medical diagnosis and Astronomy.

Test: Bragg’s Law - Question 5

The diffraction pattern of copper metal was measured with X-ray radiation of wavelength of 1.315 Å. The first order Bragg diffraction peak was found at an angle 2θ of 60°. The d-spacing between the diffracting planes in the copper metal is

Detailed Solution for Test: Bragg’s Law - Question 5

Concept:
Bragg’s Law states that when an X-ray beam are allowed to scatter through a crystalline surface, its angle of incidence θ will reflect with some angle of scattering θ. And when the path difference d is equal to a whole number n, of wavelength.
It can be represented as
2dsinθ  =nλ
Calculation:
Given:

λ = 1.315 Å, n = 1,
2θ = 60°
θ = 30°
From Bragg’s law:
2d sin θ = nλ
2 × d × sin 30° = 1 × 1.315

Test: Bragg’s Law - Question 6

A beam of X-rays is constructively scattered in second order from the surface of the crystal at an angle of 30° and the spacing between layers of atoms in NACl crystal is 4.5 × 10-10 m. The wavelength of X-rays is ______ 

Detailed Solution for Test: Bragg’s Law - Question 6

Concept:
Bragg’s Law:
Bragg's Law relates the angle θ (at which there is a maximum in diffracted intensity ) to the wavelength of X-rays and the interlayer distance d between the planes of atoms/ions/molecules in the lattice. Condition for constructive interference (maxima in the reflected).
nλ = 2d sinθ
λ = Wavelength of the X-ray
d = distance of crystal layers
θ = incident angle (the angle between the incident plane and the scattering plane)
n = integer (order)
Calculation:
Given:
d = 4.5 × 10-10 m, θ = 30°, n = 2
We know that,
nλ = 2d sin θ
∴ 2λ = 2 × 4.5 × 10-10 × sin 30°
λ = 2.25 × 10-10 m

Test: Bragg’s Law - Question 7

X-ray crystallography is not used to find the physical properties of which of the following?

Detailed Solution for Test: Bragg’s Law - Question 7

CONCEPT:

  • X-ray crystallography: It is the technique of determining the molecular and atomic structure of a crystal.
  • X-ray beams are used to determine the structure of the crystal.
  • The crystal diffracts the X-ray beams and the intensities and angle of diffraction of these diffracted beams is studied to estimate the structure of the crystal.

EXPLANATION:

  • The diffraction of light is used in X-ray crystallography.
  • X-rays diffracted from the crystal are studied to determine the structure of the crystal.
  • X-ray crystallography is not used to find the physical properties of liquids.
Test: Bragg’s Law - Question 8

The characteristic of light used in X-Ray crystallography is-

Detailed Solution for Test: Bragg’s Law - Question 8

CONCEPT:

  • X-ray crystallography: It is the technique of determining the molecular and atomic structure of a crystal.
  • X-ray beams are used to determine the structure of the crystal.
  • The crystal diffracts the X-ray beams and the intensities and angle of diffraction of these diffracted beams are studied to estimate the structure of the crystal.

EXPLANATION:

  • The diffraction of light is used in X-ray crystallography.
  • X-rays diffracted from the crystal are studied to determine the structure of the crystal.
Test: Bragg’s Law - Question 9

Two lines, A and B, of an X-ray beam, give second-order reflection maximum at a glancing angle of 60° and third-order reflection maximum at an angle of 30° respectively from the face of the same crystal. If the wavelength of line B is 0.25 nm, then the wavelength of line A will be:

Detailed Solution for Test: Bragg’s Law - Question 9

Concept:
In X-ray scattering
nλ = 2d sin θ

where d = distance between atomic planes, θ = glancing angle, λ = wavelength
Calculation:
Given:
For line A:
n1 = 2, θ1 = 60o, λ1 = ?
For line B;
n2 = 3, θ2 = 30o, λ2 = 0.25 nm
Distance between atomic planes d1 and d2 is equal for same crystal i.e.;
d1 = d2 = d
From equation (1);


On solving we'll get;
λ1 = 0.649 nm ≈ 0.65 nm

Test: Bragg’s Law - Question 10

X-rays are _______.

Detailed Solution for Test: Bragg’s Law - Question 10

X-rays:

  • X-rays are electromagnetic radiations.
  • They are high-energy radiations that are not produced by the movement of charges.
  • This radiation has a high penetrating nature and is therefore used in scanning.
  • Similar to heat radiations, these rays are also radiations.
  • The small number of X-rays are accompanied by the sunlight that reaches the Earth.
  • Electric fields are produced by charged particles and these fields interact with charged particles.
  • Electrically charged particle motion produces a magnetic field.
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