A beam of electrons of energy E scatters from
a target having atomic spacing of 1 $\mathop A\limits^o$. The first
maximum intensity occurs at $\theta$ = 60o. Then E (in
eV) is ______.
(Planck constant h = 6.64 × 10–34 Js,
1 eV =
1.6 × 10–19 J, electron
mass m = 9.1 × 10–31 kg)
Answer (integer)
50
Solution
Given d = 1 $\mathop A\limits^o$
<br><br>For first maxima, $\theta$ = 60<sup>o</sup>
<br><br>$\therefore$ $\theta$<sub>1</sub> = 90 - ${\theta \over 2}$
<br><br>= $90 - {{60} \over 2}$ = 60<sup>o</sup>
<br><br>and $2d\sin \theta = \lambda = {h \over {\sqrt {2mE} }}$<br><br>$\Rightarrow$ $$2 \times {10^{ - 10}} \times {{\sqrt 3 } \over 2} = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {2mE} }}$$<br><br>$\Rightarrow$ $$E = {1 \over 2} \times {{6.64 \times {{10}^{ - 48}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times 1.6 \times {{10}^{ - 19}}}} = 50.47$$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Davisson-Germer Experiment
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