A) \[{{\lambda }_{p}}={{\lambda }_{n}}>{{\lambda }_{e}}>{{\lambda }_{a}}\]
B) \[{{\lambda }_{a}}<{{\lambda }_{p}}={{\lambda }_{n}}>{{\lambda }_{e}}\]
C) \[{{\lambda }_{e}}<{{\lambda }_{p}}={{\lambda }_{n}}>{{\lambda }_{a}}\]
D) \[{{\lambda }_{e}}={{\lambda }_{p}}={{\lambda }_{n}}={{\lambda }_{a}}\]
Correct Answer: B
Solution :
| Option [b] is correct. |
| Explanation: Matter waves (de-Broglie waves) According to de-Broglie, a moving material particle sometimes acts as a wave and sometimes as a particle. |
| De- Broglie wavelength: \[{{\lambda }_{d}}=\frac{h}{p}\] |
| \[{{\operatorname{E}}_{p}}={{E}_{n}}={{E}_{e}}={{E}_{x}}\] |
| \[K.\operatorname{E}.=K=\frac{1}{2}m{{v}^{2}}\] |
| \[2K=m{{v}^{2}}\] |
| \[2Km={{m}^{2}}{{v}^{2}}\] |
| \[2mK={{p}^{2}}\] |
| \[\sqrt{2mk=p}\] |
| \[\therefore {{\lambda }_{d}}\frac{h}{p}\] |
| \[{{\lambda }_{d}}=\frac{h}{\sqrt{2m\operatorname{K}}}\] |
| or \[{{\lambda }_{d}}=\frac{h}{\sqrt{2m\operatorname{K}}}\][as h and E(K.E.)is constt.] |
| \[\therefore \lambda \propto \frac{1}{\sqrt{m}}\] |
| \[{{m}_{a}}>{{m}_{p}}={{m}_{n}}>{{m}_{e}}\] |
| \[\therefore {{\lambda }_{a}}<{{\lambda }_{p}}={{\lambda }_{n}}<{{\lambda }_{e}}\] |
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