Question 171: Atoms & Nuclei

Question

If the wavelength of the first member of Lyman series of hydrogen is $$\lambda$$. The wavelength of the second member will be

Accepted Answer

$$\begin{aligned}
& \frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}ight] ...... 	ext{(i)}\
& \frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}ight] ...... 	ext{(ii)}
\end{aligned}$$
On dividing (i) & (ii)
$$\lambda^{\prime}=\frac{27}{32} \lambda$$

Answer

$$\frac{27}{5} \lambda$$

Answer

$$\frac{5}{27} \lambda$$

Answer

$$\frac{27}{32} \lambda$$

Answer

$$\frac{32}{27} \lambda$$

Atoms & Nuclei

If the wavelength of the first member of Lyman series of hydrogen is $\lambda$. The wavelength of the second member will be

Choose the correct answer: