Ncert Exercises & Solutions

A neutron beam of energy E scatters from atoms on a surface with a spacing d=0.1 nm. The first maximum of intensity in the reflected beam occurs at $θ = 30 ° .$ What is the kinetic energy E of the beam in eV?

Hint: The momentum of the particles depends on the wavelength of the beam.

Step 1: Find the wavelength of the beam.

Given, d=0.1 nm, $θ = 30 ° and n = 1$

Now, according to Bragg's law

$\begin{aligned} 2 dsinθ & = nλ \Rightarrow 2 \times 0 .1 \times \sin 30 ° = λ \end{aligned}$
$\begin{aligned} \Rightarrow λ & = 0 .1 nm = 10^{- 10} m \end{aligned}$
$Step 2 : Find the momentum of the particles .$
$\begin{aligned} Now, λ & = \frac{h}{mv} = \frac{h}{p} \end{aligned}$
$\begin{aligned} \Rightarrow p = \frac{h}{λ} & = \frac{6 .62 \times 10^{- 34}}{10^{- 10}} \Rightarrow = 6 .62 \times 10^{- 24} kg - m / s \end{aligned}$
$Step 3 : Find the kinetic energy of the particles .$
$\begin{aligned} Now, KE & = \frac{1}{2} {mv}^{2} = \frac{1}{2} \frac{m^{2} v^{2}}{m} = \frac{1}{2} \frac{p^{2}}{m} = \frac{1}{2} \times \frac{{(6 .62 \times 10^{- 24})}^{2}}{1 .67 \times 10^{- 27}} J = 0 .21 eV \end{aligned} \begin{array}{l} \end{array}$

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