A) 5 : 1
B) 4 : 1
C) 1 : 5
D) 1 : 4
Correct Answer: D
Solution :
[d] From the expression of Bohr's theory, we know that \[{{m}_{e}}{{v}_{1}}{{r}_{1}}={{n}_{1}}\frac{h}{2\pi }\] \[\And \,{{m}_{e}}{{v}_{2}}{{r}_{2}}={{n}_{2}}\frac{h}{2\pi }\] \[\frac{{{m}_{e}}{{v}_{1}}{{r}_{1}}}{{{m}_{e}}{{v}_{2}}{{r}_{2}}}=\frac{{{n}_{1}}}{{{n}_{2}}}\frac{h}{2\pi }\times \frac{2\pi }{h}\] Given, \[{{r}_{1}}=5{{r}_{2}},\,{{n}_{1}}=5,{{n}_{2}}=4\] \[\frac{{{m}_{e}}\times {{v}_{1}}\times 5{{r}_{2}}\text{ }}{{{m}_{e}}\times {{v}_{2}}\times {{r}_{2}}}=\frac{5}{4}\] \[\Rightarrow \frac{{{v}_{1}}}{{{v}_{2}}}=\frac{5}{4\times 5}=\frac{1}{4}=1 : 4\]
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