The de-Broglie wavelength of a particle having kinetic energy E is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?
Solution
$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$, $mv = \sqrt {2mE}$<br><br>$\lambda \propto {1 \over {\sqrt E }}$<br><br>$${{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{E_1}} \over {{E_2}}}} = {3 \over 4}$$, ${\lambda _2} = 0.75{\lambda _1}$<br><br>${{{E_1}} \over {{E_2}}} = {\left( {{3 \over 4}} \right)^2}$<br><br>${E_2} = {{16} \over 9}{E_1} = {{16} \over 9}E$ (E<sub>1</sub> = E)<br><br>Extra energy given = ${{16} \over 9}E - E = {7 \over 9}E$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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