A) equal to half the circumference of the first orbit
B) equal to one fourth the circumference of the first orbit
C) equal to the circumference of the first orbit
D) equal to twice the circumference of the first orbit
Correct Answer: C
Solution :
de-Broglie wavelength \[\lambda =\frac{h}{p}\] \[\therefore \] \[L=\frac{nh}{2\pi }\]for \[{{\text{H}}_{\text{2}}}\]atom (Bohr theory.) \[L=\]angular momentum \[n=\]principal quantum number \[\therefore \] \[L=m\upsilon \,\,\,{{r}_{n}}=p{{r}_{n}}\] \[{{r}_{n}}=\]radius of orbit \[p{{r}_{n}}=\frac{hn}{2\pi }\] \[\frac{h}{p}=\lambda \Rightarrow \frac{2\pi {{r}_{n}}}{n}\] for \[n=1\]first Bohr's orbit \[\lambda =2\pi \,{{r}_{1}}=\]circumference of first Bohr orbits.
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