Question

If ${\lambda }_{L1},{\lambda }_{L2}$ and ${\lambda }_{L3}$ are wavelengths of $1^{st},2^{nd}$ and $3^{rd}$ lines of Lyman series ${\lambda }_{B1},{\lambda }_{B2}$ and ${\lambda }_{B3}$ are wavelengths of $1^{st},2^{nd}$ and $3^{rd}$ lines of Balmer series while ${\lambda }_{P1}$ and ${\lambda }_{P2}$ are wavelengths of $1^{st},2^{nd}$ lines of Paschen series of hydrogen then which of the following relations is/are correct?

• A. $\frac{1}{{\lambda }_{L3}}=\frac{1}{{\lambda }_{P1}}+\frac{1}{{\lambda }_{B1}}+\frac{1}{{\lambda }_{L1}}$
• B. $\frac{1}{{\lambda }_{B3}}=\frac{1}{{\lambda }_{P2}}+\frac{1}{{\lambda }_{B1}}$
• C. $\frac{1}{{\lambda }_{B3}}=\frac{1}{{\lambda }_{B2}}+\frac{1}{{\lambda }_{B1}}$
• D. $\frac{1}{{\lambda }_{B2}}=\frac{1}{{\lambda }_{B1}}+\frac{1}{{\lambda }_{P1}}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\frac{1}{{\lambda }_{L3}}=\frac{1}{{\lambda }_{P1}}+\frac{1}{{\lambda }_{B1}}+\frac{1}{{\lambda }_{L1}}$
B $\frac{1}{{\lambda }_{B3}}=\frac{1}{{\lambda }_{P2}}+\frac{1}{{\lambda }_{B1}}$
D $\frac{1}{{\lambda }_{B2}}=\frac{1}{{\lambda }_{B1}}+\frac{1}{{\lambda }_{P1}}$


$\frac{1}{{\lambda }_{L1}}=R_H(\frac{1}{1^2}-\frac{1}{2^2}),\frac{1}{{\lambda }_{L2}}=R_H(\frac{1}{1^2}-\frac{1}{3^2}),\frac{1}{{\lambda }_{L3}}=R_H(\frac{1}{1^2}-\frac{1}{4^2})$
$\frac{1}{{\lambda }_{B1}}=R_H(\frac{1}{2^2}-\frac{1}{3^2}),\frac{1}{{\lambda }_{B2}}=R_H(\frac{1}{2^2}-\frac{1}{4^2}),\frac{1}{{\lambda }_{B3}}=R_H(\frac{1}{2^2}-\frac{1}{5^2})$
$\frac{1}{{\lambda }_{P1}}=R_H(\frac{1}{3^2}-\frac{1}{4^2}),\frac{1}{{\lambda }_{P2}}=R_H(\frac{1}{3^2}-\frac{1}{5^2})$
(a) $\frac{1}{{\lambda }_{P1}}+\frac{1}{{\lambda }_{B1}}+\frac{1}{{\lambda }_{L1}}=R_H(1-\frac{1}{4^2})=\frac{1}{{\lambda }_{L3}}$
(b) $\frac{1}{{\lambda }_{P2}}+\frac{1}{{\lambda }_{B1}}=R_H(\frac{1}{2^2}-\frac{1}{5^2})=\frac{1}{{\lambda }_{B3}}$
(c) $\frac{1}{{\lambda }_{B2}}+\frac{1}{{\lambda }_{B1}}=(\frac{2}{2^2}-\frac{1}{4^2}-\frac{1}{3^2})$
(d) $\frac{1}{{\lambda }_{B1}}+\frac{1}{{\lambda }_{P1}}=(\frac{1}{2^2}-\frac{1}{4^2})=\frac{1}{{\lambda }_{B2}}$