A) \[{{\lambda }_{1}}<{{\lambda }_{2}}<{{\lambda }_{3}}\]
B) \[{{\lambda }_{1}}>{{\lambda }_{2}}>{{\lambda }_{3}}\]
C) \[{{\lambda }_{1}}<{{\lambda }_{2}}>{{\lambda }_{3}}\]
D) \[{{\lambda }_{1}}>{{\lambda }_{2}}<{{\lambda }_{3}}\]
Correct Answer: B
Solution :
The de-Broglie wavelength is given as \[\lambda =\frac{h}{\sqrt{2mqV}}\] Here, h and V are constants so, \[\lambda \propto \frac{1}{\sqrt{mq}}\] \[\therefore \]\[{{\lambda }_{1}}:{{\lambda }_{2}}:{{\lambda }_{3}}=\frac{1}{\sqrt{{{m}_{1}}{{q}_{1}}}}:\frac{1}{\sqrt{{{m}_{2}}{{q}_{2}}}}:\frac{1}{\sqrt{{{m}_{3}}{{q}_{3}}}}\] \[=\frac{1}{\sqrt{\frac{{{m}_{p}}}{1840}\times e}}:\frac{1}{\sqrt{{{m}_{p}}\times e}}:\frac{1}{\sqrt{4{{m}_{p}}\times 2e}}\] \[=\sqrt{1840}:1:\frac{1}{2\sqrt{2}}\] So \[{{\lambda }_{1}}>{{\lambda }_{2}}>{{\lambda }_{3}}\]
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