Ncert Exercises & Solutions

Question 11.31:

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 I Å, which is of the order of inter-atomic spacing in the lattice) (m_e= 9.11 x 10^-31 kg).

Hint: $E=\frac{p^2}{2m}$

Step 1: Find the energy of an electron probe.
$E=\frac{p^2}{2m}\\ \text{According to de broglie hypothesis}\\ \lambda=\frac{h}{p}\\ p=\frac{h}{\lambda}\\ \Rightarrow E=\frac{h^2}{2\lambda^2 m_e}$

$\begin{aligned} \begin{matrix} = \frac{{(6 .6 \times 10^{- 34})}^{2}}{2 \times {(10^{- 10})}^{2} \times 9 .11 \times 10^{- 31}} = 2 .39 \times 10^{- 17} J \\ = \frac{2 .39 \times 10^{- 17}}{1 .6 \times 10^{- 19}} = 149 .375 eV \end{matrix} \end{aligned}$

Step 2: Find the energy of a photon.
$E=\frac{hc}{\lambda e}~eV\\= \frac{6.6\times10^{-34}\times3\times10^8}{10^{-10}\times1.6\times10^{-19}}\\=12.375 ~KeV$

Hence energy of x-rays is greater.

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