Question

In a certain electronic transition in Hydrogen atom from an initial state to a final state, the difference of orbit radius is $8$ times the first Bohr radius. Which transition will satisfy the given condition?

• A. $7⟶1$
• B. $6⟶1$
• C. $5⟶1$
• D. $3⟶1$

Answer 1

Answer: (D)

$3⟶1$

Radius of $n^{th}$ bohr orbit $\Rightarrow r_n=\frac{n^2{h}^2}{4{\pi }^{2}mZ{e}^2}$

Given difference of bohr radius from initial to a final state. $d$ is $8$ times the first bohr radius.

Let initial state be $n_1$ and final state be $n_2$

$r_{n_2}=\frac{n_2^2{h}^2}{4{\pi }^{2}mZ{e}^2}....\left(i\right)$

$r_{n_1}=\frac{n_1^2{h}^2}{4{\pi }^{2}mZ{e}^2}....\left(ii\right)$

$r_{n_2}-r_{n_1}=\frac{8h^2}{4{\pi }^{2}mZ{e}^2}$

From (i) and (ii),

$n_2^2-n_1^2=8$

So, only tranisition $3\to 1$ satisfied the given condition.