A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :
Solution
${\lambda _p} = {h \over {{P_p}}}$<br><br>${\lambda _e} = {h \over {{P_e}}}$<br><br>$\because$ ${\lambda _p} = {\lambda _e}$<br><br>$\Rightarrow {P_p} = {P_e}$<br><br>${(K)_p} = {{P_p^2} \over {2{m_p}}}$<br><br>${(K)_e} = {{P_e^2} \over {2{m_e}}}$<br><br>K<sub>p</sub> < K<sub>e</sub> as m<sub>p</sub> > m<sub>e</sub><br><br>Option (a)
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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