A) \[{{\lambda }_{CM}}={{\lambda }_{1}}={{\lambda }_{2}}\]
B) \[\frac{1}{{{\lambda }_{CM}}}=\frac{1}{{{\lambda }_{1}}}+\frac{1}{{{\lambda }_{2}}}\]
C) \[{{\lambda }_{CM}}=\frac{2{{\lambda }_{1}}{{\lambda }_{2}}}{\sqrt{{{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}}}\]
D) \[{{\lambda }_{CM}}=\left( \frac{{{\lambda }_{1}}+{{\lambda }_{2}}}{2} \right)\]
Correct Answer: B
Solution :
| [b] Since |
| \[\lambda =\frac{h}{mv}\] |
| \[v=\frac{h}{m\lambda }\] |
| Let \[{{v}_{1}}\] and \[{{v}_{2}}\] are the speeds of electrons |
| \[{{v}_{cm}}=\frac{{{v}_{1}}+{{v}_{2}}}{2}\] |
| \[\frac{h}{2m{{\lambda }_{cm}}}=\frac{1}{2}(\frac{h}{m{{\lambda }_{l}}}+\frac{h}{m{{\lambda }_{2}}})\] |
| \[\frac{1}{{{\lambda }_{cm}}}=\frac{1}{{{\lambda }_{1}}}+\frac{1}{{{\lambda }_{2}}}\] |
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