A particle moving with kinetic energy E has de Broglie wavelength $\lambda$. If energy $\Delta$E is added to its energy, the wavelength become $\lambda$/2. Value of $\Delta$E, is :
Solution
$\lambda = {h \over {\sqrt {2mE} }}$
<br><br>Also, ${h \over {\sqrt {2m\left( {E + \Delta E} \right)} }}$ = ${\lambda \over 2}$
<br><br>$\therefore$ ${{E + \Delta E} \over E} = 4$
<br><br>$\Rightarrow$ $\Delta$E = 3E
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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