$\begin {array} {1 1} (a)\;X-ray\: probe & \quad (b)\;Electron\: probe \\ (c)\;Both\: have\: equal\: energies & \quad (d)\;Can’t \: be\: determined \end {array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Ans : (a)

$ \lambda = 1A^{\circ}$

$E=\large\frac{1}{2} m_ev^2$

$ \Rightarrow \: m_ev = \sqrt 2Em_e\: \: \: \: \: where,\: v=$ velocity of electron; p(momentum) = $m_e$

So,$ \lambda = \large\frac{h}{p} = \large\frac{h}{m_ev} = \large\frac{h}{\sqrt2 Em_e}$

So, $E = \large\frac{h^2}{2 \lambda^2m_e}$

$= \large\frac{(6.6 \times 10^{-34} )^2}{ [2 \times (10^{-10} )^2 \times 9.11 \times 10^{-31} ]}$

$= \large\frac{2.39 \times 10^{-17} J}{1.6 \times 10^{-19} } = 149.375 \: eV$

$E’ $ = energy of a photon = $\large\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{ (10^{-10} \times 1.6 \times 10^{-19 } )}$

$= 12.375 \times 10^3\: eV = 12.375\: keV$

So, $E’>E$, i.e, X-ray probe has greater energy than electron probe.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...