Which transition occurs when light with a wavelength of 434 nm is emitted by a hydrogen atom? Group of answer choicesThe electron falls from n=5 to n=2The electron falls from n=4 to n=2The electron jumps from n=2 to n=4The electron jumps from n=2 to n=5

To determine which transition occurs when light with a wavelength of 434 nm is emitted by a hydrogen atom, we can use the Rydberg formula for hydrogen emission:

$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

where:

• $\lambda$ is the wavelength of the emitted light,
• $R_H$ is the Rydberg constant ($1.097 \times 10^7 \, \text{m}^{-1}$),
• $n_1$ and $n_2$ are the principal quantum numbers of the electron's initial and final states, respectively, with $n_2 > n_1$.

Given $\lambda = 434 \, \text{nm} = 434 \times 10^{-9} \, \text{m}$, we can rearrange the Rydberg formula to solve for the transition:

$\frac{1}{434 \times 10^{-9} \, \text{m}} = 1.097 \times 10^7 \, \text{m}^{-1} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

First, calculate the left side of the equation:

$\frac{1}{434 \times 10^{-9} \, \text{m}} = 2.304 \times 10^6 \, \text{m}^{-1}$

Now, set up the equation:

$2.304 \times 10^6 \, \text{m}^{-1} = 1.097 \times 10^7 \, \text{m}^{-1} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Divide both sides by $1.097 \times 10^7 \, \text{m}^{-1}$:

$\frac{2.304 \times 10^6}{1.097 \times 10^7} = \frac{1}{n_1^2} - \frac{1}{n_2^2}$

$0.210 = \frac{1}{n_1^2} - \frac{1}{n_2^2}$

We know that the Balmer series involves transitions where $n_1 = 2$. Therefore:

[ 0.210 = \frac{1}{2^2} - \frac{1}{n_2^2}