An electron in a hydrogen atom undergoes a transition from an orbit with quantum number to another with quantum number . and are respectively the initial and final potential energies of the electron. If then the smallest possible is

Question

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $n_{i}$ to another with quantum number $n_{f}$ . $V_{i}$ and $V_{f}$ are respectively the initial and final potential energies of the electron. If $\frac{V_{i}}{V_{f}} = 6.25,$ then the smallest possible $n_{f}$ is [2017]

Smart Achievers · Accepted Answer

(5)

Here, $\frac{V_{i}}{V_{f}} = \frac{\frac{- 27.2}{n_{f}^{2}}}{\frac{- 27.2}{n_{i}^{2}}} = \frac{n_{f}^{2}}{n_{i}^{2}} = 6.25$

$∴$ $\frac{n_{f}}{n_{i}} = 2.5 = \frac{5}{2}$

$∴$ Smallest possible $n_{f} = 5$

Solution

Ans.
(5)

Here, $\frac{V_{i}}{V_{f}} = \frac{\frac{- 27.2}{n_{f}^{2}}}{\frac{- 27.2}{n_{i}^{2}}} = \frac{n_{f}^{2}}{n_{i}^{2}} = 6.25$

$∴$ $\frac{n_{f}}{n_{i}} = 2.5 = \frac{5}{2}$

$∴$ Smallest possible $n_{f} = 5$

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Solution

Q. An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number nf. Vi and Vf are respectively the initial and final potential energies of the electron. If ViVf=6.25, then the smallest possible nf is [2017]

Ans. (5) Here, ViVf=-27.2nf2-27.2ni2=nf2ni2=6.25 ∴ nfni=2.5=52 ∴ Smallest possible nf=5

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Ans.
(5)

Here, $\frac{V_{i}}{V_{f}} = \frac{\frac{- 27.2}{n_{f}^{2}}}{\frac{- 27.2}{n_{i}^{2}}} = \frac{n_{f}^{2}}{n_{i}^{2}} = 6.25$

$∴$ $\frac{n_{f}}{n_{i}} = 2.5 = \frac{5}{2}$

$∴$ Smallest possible $n_{f} = 5$