Question

In a hydrogen atom, an electronic transition takes place from an initial state (1) to a final state (2). The difference in the orbit radius $(r_1-r_2)$ is 24 times the first Bohr radius. Identify the transition.

• A. $5\to 1$
• B. $25\to 1$
• C. $8\to 3$
• D. $7\to 5$

Answer 1

Answer: (A), (D)

$7\to 5$

Let:
initial state (1) have orbit number $=n_1$
final state (2) have orbit number $=n_2$

Bohr's radius for $n^{th}$ orbit, $\left(r_n\right)=0.529\left[\frac{n^2}{Z}\right]\stackrel{\circ }{A}$
where, $n=\text{Orbit number},Z=\text{Atomic number}$

Given: Difference in orbit radius $=r_1-r_2=24\times \text{radius of 1st Bohr orbit}$

As the element taken is hydrogen, hence $Z=1$
For 1st Bohr orbit, $n=1$
Radius for first Bohr orbit, $r_0=0.529\stackrel{\circ }{A}$

Hence, $r_{n_1}-r_{n_2}=24\times r_0=12.7\stackrel{\circ }{A}$

For $n_1$,
$r_{n1}=0.529\left[\frac{n_1^2}{Z}\right]=0.529n_1^2...\left(i\right)$

For $n_2$,
$r_{n2}=0.529\left[\frac{n_2^2}{Z}\right]=0.529n_2^2...\left(ii\right)$

Taking the difference of equations 1 and 2, we get
$r_{n_1}-r_{n_2}=\left(0.529n_1^2\right)-\left(0.529n_2^2\right)$
$=0.529(n_1^2-n_2^2)$
$\Rightarrow 12.7=0.529(n_1^2-n_2^2)$

$\Rightarrow (n_1^2-n_2^2)=\frac{12.7}{0.529}=24$
$\Rightarrow (n_1-n_2)(n_1+n_2)=24$

Now there are 3 possibilities:
1. $(n_1-n_2)(n_1+n_2)=12\times 2$
2. $(n_1-n_2)(n_1+n_2)=6\times 4$
3. $(n_1-n_2)(n_1+n_2)=8\times 3$

For case (1),
Let $(n_1-n_2)=2$ and $(n_1+n_2)=12$
Upon solving, we get
$n_1=7$ and $n_2=5$
Hence the transition would be $7\to 5$

For case (2),
Let $(n_1-n_2)=4$ and $(n_1+n_2)=6$
Upon solving, we get
$n_1=5$ and $n_2=1$
Hence the transition would be $5\to 1$

For case (3),
Let $(n_1-n_2)=3$ and $(n_1+n_2)=8$
Upon solving, we get
$n_1$ = 5.5 and $n_2$ = 2.5 which are not possible

Only $7\to 5$ and $5\to 1$ transitions are the correct answers.