A) \[1.67\overset{o}{\mathop{A}}\,\]
B) \[3.33\overset{o}{\mathop{A}}\,\]
C) \[1.06\overset{o}{\mathop{A}}\,\]
D) \[0.53\overset{o}{\mathop{A}}\,\]
Correct Answer: B
Solution :
According to Bohr’s quantitation of angular momentum \[mvr=\frac{nh}{2\pi }\] or \[\frac{h}{mv}=\frac{2\pi r}{n}\] ... (i) de- Broglie wavelength \[\lambda =\frac{h}{mv}\] ... (ii) From Eqs. (i) and (ii), we get Wavelength \[\lambda =\frac{2\pi r}{n}\] \[=\frac{2\times \pi \times 0.53\,\overset{o}{\mathop{A}}\,}{1}\] \[=3.33\overset{o}{\mathop{A}}\,\]
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