Question

Last line of Lyman series for H-atom has wavelength $\lambda_{1} \mathring{A}$ The $2^{\text {nd }}$ line of Balmer series has wavelength $\lambda_{2} \mathring{A},$ then:

• A. $\frac{16}{\lambda_{1}}=\frac{9}{\lambda_{2}}$
• B. $\frac{16}{\lambda_{2}}=\frac{3}{\lambda_{1}}$
• C. $\frac{4}{\lambda_{1}}=\frac{1}{\lambda_{2}}$
• D. $\frac{16}{\lambda_{1}}=\frac{3}{\lambda_{2}}$

Answer 1

Answer: (B)

Last line of lyman series refers to transition $\infty \rightarrow 1$

$\frac{1}{\lambda_{1}}= R \left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right)$

$\frac{1}{\lambda_{1}}= R ...(1)$

$2^{nd}$ line of Balmer series refers to transition $4 \rightarrow 2$

$\frac{1}{\lambda_{2}}= R \left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$

$\frac{1}{\lambda_{2}}= R \left(\frac{1}{4}-\frac{1}{16}\right)$

$\frac{1}{\lambda_{2}}= R \times \frac{3}{16}...(2)$

$\frac{\frac{1}{\lambda_{1}}}{\frac{1}{\lambda_{2}}}$

$=\frac{ R }{\frac{3 R }{16}}=\frac{ R }{1} \times \frac{16}{3 R }$

$=\frac{\lambda_{2}}{\lambda_{1}}=\frac{16}{3}$