A) Electron
B) Alpha particle
C) Proton
D) Neutron
Correct Answer: A
Solution :
de-Broglie wavelength \[\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}}\] E is same for all, so \[\lambda \propto \frac{1}{\sqrt{m}}\] Hence, de-Broglie wavelength will be maximum for particle with lesser mass. Mass of the given particles in increasing order are given as \[{{m}_{e}}<{{m}_{p}}<{{m}_{n}}<{{m}_{\alpha }}\] Thus, de-Broglie wavelength will be maximum for electron.
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