Crystal diffraction experiments can be performed using 𝑋−rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to $1 \circ A$ , which is of the order of inter-atomic spacing in the lattice) $(m_{e} = 9.11 \times 10^{- 31} kg)$

Open in App

Solution

Step 1 : Generate the formula for momentum.
Formula used: p = mv
Given:
Wavelength of the probe,
$λ = 1 \circ A = 10^{- 10} m$
Mass of an electron,
$m_{e} = 9.11 \times 10^{- 31} kg$
Planck's constant,
$h = 6.6 \times 10^{- 34} J s$
Charge on an electron,
$e = 1.6 \times 10^{- 19} C$
The kinetic energy of the electron is given as:
$K = \frac{1}{2} m_{e} v^{2}$
It can be written as,
$m_{e} v = \sqrt{2 K m_{e}}$
Step 2: Generate the formula for de Broglie wavelength.
Formula used: $λ = \frac{h}{p}$
Also, according to the de Broglie principle, the de Broglie wavelength is given as:
$λ = \frac{h}{p} . . . . . (i i)$
From (1) &(2) equation, $λ = \frac{h}{m_{e} v}$
$λ = \frac{h}{\sqrt{2 K m_{e}}}$
Step 3: Find the electron’s energy (K).
$K = \frac{h^{2}}{2 λ^{2} m_{e}}$
By putting all the given values,
$K = \frac{(6.626 \times 10^{- 34})^{2}}{2 \times (10^{- 10})^{2} \times 9.1 \times 10^{- 31}}$
$K = 2.4 \times 10^{- 17} J$
To convert 𝐸 into 𝑒𝑣
$K = \frac{2.4 \times 10^{- 17}}{1.6 \times 10^{- 19}} e V$
$K = 150 e V$
Step 4: Find the photon’s energy $(K^{')}$
. $K^{'} = \frac{h c}{λ e}$
By putting all the given values,
$K^{'} = \frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{10^{- 10} \times 1.6 \times 10^{- 19}}$
$K^{'} = 12.4 keV$
Hence, K' > K for the same wavelength.
Final Answer : $K = 150 e V$
$K^{'} = 12.4 keV$

Suggest Corrections

Similar questions

Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (m_e= 9.11 × 10⁻³¹ kg).

Q. (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain.(

m_{n} = 1.675 \times 10^{27}

kg)
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (

27^{o} C

). Hence, explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Exercise 11.31:

A∘A∘ which is of the order of inter-atomic spacing in the lattice) (

m_{e} = 9.11 \times 10^{- 31} k g

)

Q. What will be the wavelength of the X-rays which give a diffraction angle

2 θ

equal to

{16.80}^{o}

for a crystal, if the inter planner distance in the crystal is 0.200 nm and only first order diffraction is observed

(sin {8.40}^{o} = 0.146) .

Q. An X-ray beam of wavelength

1.0 ˚ A

is incident on a crystal of lattice spacing

2.8 ˚ A

. The value of Bragg angle for first order of diffraction is :

The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10⁻¹⁵ m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)

Join BYJU'S Learning Program

Explore more

Creating AM

Standard XII Physics

Join BYJU'S Learning Program