A) \[6.6\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[1.2\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[0.133\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[0.41\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: D
Solution :
Using the relation, \[\lambda =\frac{12375}{eV}{{\overset{\text{o}}{\mathop{\text{A}}}\,}_{unit}}\] Given, \[V=30\text{ }kV\text{ }=30000\] \[\lambda =\frac{12375}{30000}=0.41\overset{\text{o}}{\mathop{\text{A}}}\,\] For hydrogen atom \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] According to de Broglie \[p=\frac{h}{\lambda }\] Momentum \[p=hR\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[p=6.6\times {{10}^{-34}}\times 1.0974\times {{10}^{7}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] \[p=6.8\times {{10}^{-27}}Nm\]
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