Question 308: Atoms & Nuclei

Question

An electron in the hydrogen atom initially in the fourth excited state makes a transition to $\mathrm{n}^{	ext {th }}$ energy state by emitting a photon of energy 2.86 eV . The integer value of n will be__________.

Accepted Answer

To find the integer value of $ n $ for which an electron transitions from the fourth excited state in a hydrogen atom, thus emitting a photon with an energy of 2.86 eV, we can follow these steps:
We use the formula for the energy difference associated with electron transitions in a hydrogen atom:
$    E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)    $
However, in this question, it seems there's a typo in the original explanation, as both indices are given as $\mathrm{n}_1$. To correct it, we should use this formula:
$    E = 13.6 \left( \frac{1}{n^2} - \frac{1}{5^2} \right)    $
where $ n_1 = 5 $ (the fifth energy level or fourth excited state) and $ n_2 = n $ (the state to which the electron transitions).
Given the photon's energy is 2.86 eV, set up the equation:
$    2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{25} \right)    $
Solve for $\frac{1}{n^2}$:
$    \frac{1}{n^2} = \frac{2.86}{13.6} + \frac{1}{25}    $
Calculate the value:
$    \frac{1}{n^2} = 0.21 + 0.04 = 0.25    $
Find $ n^2 $:
$    n^2 = \frac{1}{0.25} = 4    $
Consequently:
$    n = \sqrt{4} = 2    $
Thus, the electron transitions to the $ n = 2 $ energy state.

Atoms & Nuclei

An electron in the hydrogen atom initially in the fourth excited state makes a transition to \mathrm{n}^{\text {th }} energy state by emitting a photon of energy 2.86 eV . The integer value of n will be__________.

Integer / Numeric Answer Type