Calculate the wavelength of the light emitted when an electron in a hydrogen atom makes the transition from n=4 to n=2 using the Bohr equation.

• A. 656 nm
• B. 486 nm
• C. 486.1 nm
• D. 410 nm

Answer: (C)

Verified step by step guidance

1. Start by understanding the Bohr model, which describes the electron transitions in a hydrogen atom. The energy difference between two levels can be calculated using the formula: $E=R(\frac{1}{n^2}-\frac{1}{m^2})$, where $R$ is the Rydberg constant, $n$ and $m$ are the principal quantum numbers of the initial and final states.
2. Identify the initial and final energy levels for the transition. In this problem, the electron transitions from $n=4$ to $n=2$.
3. Calculate the energy difference using the Bohr equation: $E=R(\frac{1}{2^2}-\frac{1}{4^2})$. Substitute the values for $R$ and the quantum numbers.
4. Convert the energy difference to wavelength using the equation: $E=\frac{\mathrm{hc}}{\lambda }$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. Rearrange to solve for $\lambda$: $\lambda =\frac{\mathrm{hc}}{E}$.
5. Substitute the calculated energy difference and constants into the wavelength equation to find the wavelength of the emitted light. This will give you the wavelength in nanometers, which corresponds to one of the given options.