The relation below correctly represents the frequency of light emitted when an electron in a hydrogen atom jumps from a higher $(n_i)$ energy level to $(n_f)$ a lower energy level: $v = Rc(\frac{1}{n_f^2}-\frac{1}{n_i^2})$ Predict the values of $n_i, n_f$, respectively for the Paschen series, which will give the highest wavelength in the series

4, 3

The correct answer is Option (2) → 4, 3

For hydrogen atom, the frequency of emitted light:

$\nu = R c \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

Wavelength $\lambda = \frac{c}{\nu} \Rightarrow \lambda \propto \frac{1}{(1/n_f^2 - 1/n_i^2)}$

Paschen series: $n_f = 3$, $n_i = 4,5,6,...,\infty$

Highest wavelength corresponds to lowest frequency, which occurs when $n_i$ is just above $n_f$:

Highest wavelength → $n_i = 4$, $n_f = 3$

Answer: $n_i = 4$, $n_f = 3$