A) 22
B) 23
C) 24
D) 25
Correct Answer: D
Solution :
\[2\pi {{r}_{n}}=n\lambda \] \[\frac{1}{\lambda }=\frac{n}{2\pi {{r}_{n}}}=\frac{m{{e}^{2}}z}{2{{\varepsilon }_{0}}{{h}^{2}}n}\] ?.. (i) Also, for lyman series, \[\frac{1}{\lambda }=R{{z}^{2}}\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{n}^{2}}} \right]\] ?..(ii) From eq. (i) and (ii), n=25You need to login to perform this action.
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