Question

According to Bohr atom model, in which of the following transitions will the frequency be maximum?

• A. $n=2ton=1$
• B. $n=4ton=3$
• C. $n=5ton=4$
• D. $n=3ton=2$

Answer 1

The correct option is A

$n=2ton=1$

Explanation for the correct option:

According to Bohr's Atomic Model, minimum energy required to attain that orbit is shown in figure below

$$\begin{gathered} \mathbf{5}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }{\mathit{E}}_{\mathbf{5}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{54}\mathit{e}\mathit{V} \\ \mathbf{4}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }{\mathit{E}}_{\mathbf{4}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{85}\mathit{e}\mathit{V} \\ \mathbf{3}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }{\mathit{E}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{51}\mathit{e}\mathit{V} \\ \mathbf{2}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }{\mathit{E}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{40}\mathit{e}\mathit{V} \\ \mathbf{1}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }{\mathit{E}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{4}\mathit{e}\mathit{V} \end{gathered}$$

We know, $∆E=hf$

where, $f$ is frequency

and $∆E=13.4Z^2\left(\frac{1}{n^2}-\frac{1}{{n_1}^2}\right)$

Since, $\Delta E$ is maximum for the transition from $n=2$to $n=1$

Frequency is more for transition from $n=2$ to $n=1$.

Hence, option A is correct.