A) \[\frac{1}{\sqrt{2}},\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{\sqrt{2}},\frac{1}{2},\frac{1}{\sqrt{2}}\]
C) \[1,\sqrt{3},1\]
D) \[1,\frac{1}{\sqrt{3}},1\]
Correct Answer: B
Solution :
Let \[_{\text{5}}{{\text{Y}}^{\text{10}}}{{\text{=}}_{\text{5}}}{{\text{B}}^{\text{10}}}\],\[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] and \[\Rightarrow \] be the angles made by the line segment OP with X-axis, Y-axis and Z-axis, respectively. Given \[34=-\frac{13.6}{{{n}^{2}}}\] \[\Rightarrow \] \[{{n}^{2}}=4\] \[\Rightarrow \] \[L=\frac{nh}{2\pi }=\frac{2h}{2\pi }=\frac{h}{\pi }\] \[\because \] \[\lambda =\frac{h}{p}\] \[\therefore \] \[\lambda \alpha \frac{1}{p}\] \[\Rightarrow \] \[\frac{\Delta p}{p}=-\frac{\Delta \lambda }{\lambda }\] \[\therefore \] \[\left| \frac{\Delta p}{p} \right|=\left| \frac{\Delta \lambda }{\lambda } \right|\] \[\Rightarrow \] \[\frac{p'}{p}=\frac{0.20}{100}=\frac{1}{500}\] \[\Rightarrow \] \[{{E}_{0}}.\] Hence, direction cosines are \[{{E}_{0}}.\]i.e. \[\Delta {{I}_{C}}=1mA={{10}^{-3}}A\]
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