Question 456: Atoms & Nuclei

Question

According to Bohr's theory of hydrogen atom, the ratio of the maximum and minimum wavelength of Lyman series will be

Accepted Answer

$$\begin{aligned}
&\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}ight)\
&	ext { For Lyman series, } \mathrm{n}_1=1\
&\begin{aligned}
& \frac{1}{\lambda_{\max }}=R\left(\frac{1}{1^2}-\frac{1}{2^2}ight)=\frac{3}{4} R \
& \frac{1}{\lambda_{\min }}=R\left(\frac{1}{1^2}-\frac{1}{\infty^2}ight)=R
\end{aligned}\
&	herefore \quad \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{4}{3}
\end{aligned}$$

Answer

$3: 4$

Answer

$4: 3$

Answer

$2: 5$

Answer

$5: 2$

Atoms & Nuclei

According to Bohr's theory of hydrogen atom, the ratio of the maximum and minimum wavelength of Lyman series will be

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