Question

Match the following:
$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \left(A\right)n=6\to n=3\text{(In H- atoms)}& \left(p\right)10\text{lines in the spectrum}\\ \left(B\right)n=7\to n=3\text{(In H atoms)}& \left(q\right)\text{Spectral lines in visible region}\\ \left(C\right)n=5\to n=2\text{(In H-atoms)}& \left(r\right)\text{6 lines in the spectrum}\\ \left(D\right)n=5\to n=1\text{(In H-atoms)}& \text{(s) Spectral lines in infrared region}\\ & \text{(t) Spectral lines in U.V. region}\end{array}$

• A. $(A\to r,s)(B\to p,s);(C\to q,r,s);(D\to p,q,s,t)$
• B. $(A\to r)(B\to p,s);(C\to q,r,s);(D\to p,q,s,t)$
• C. $(A\to r,s)(B\to p,s);(C\to q,r);(D\to p,q,s,t)$
• D. $(A\to r,s)(B\to p,s);(C\to q,r,s);(D\to p,q,s)$

Answer 1

The correct option is A $(A\to r,s)(B\to p,s);(C\to q,r,s);(D\to p,q,s,t)$

(A) $6\to 3△n=3$
$\therefore \text{no. of lines}=\frac{3(3+1)}{2}=6$
All lines are in infrared region

(B) $7\to 3△n=4$
$\therefore \text{no. of lines}=\frac{4(4+1)}{2}=10$
All lines are in infrared region

(C) $5\to 2△n=3$
$\therefore \text{no. of lines}=\frac{3(3+1)}{2}=6$
Lines are in visible region and also in infrared region

(D) $5\to 1$ $△n=4$
$\therefore \text{no. of lines}=\frac{5\times 4}{2}=10$
Lines are in visible region, in Infrared region and also in U.V. region.