The first Bohr radius,r₀,is 0.0529 nm and the corresponding energy,E₀,is 13.6 eV.The wavelength of the light emitted as a hydrogen atom undergoes a transition from state n = 4 to n = 2 is A)656 nm B)486 nm C)434 nm D)410 nm E)None of these is correct.

Question

Quizplus · Accepted Answer

The wavelength of the light emitted during a transition between two energy levels in a hydrogen atom can be calculated using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, where $R$ is the Rydberg constant, $n_1$ and $n_2$ are the principal quantum numbers of the lower and upper energy levels, respectively, and $\lambda$ is the wavelength of the emitted light. For a transition from $n = 4$ to $n = 2$, the formula simplifies to calculate the wavelength of the emitted photon. Plugging in the values ($n_1 = 2$, $n_2 = 4$), and using the Rydberg constant appropriately, yields a wavelength that corresponds to 486 nm, which is characteristic of the Balmer series in the visible spectrum.

The First Bohr Radius,r0,is 0

The First Bohr Radius,r₀,is 0