The expression for longest wavelength in Paschen series (for H atom) is 144R/x. Find x. R is Rydberg’s constant.

The correct answer is 7.

To find the value of $x$ in the expression for the longest wavelength in the Paschen series for a hydrogen atom, we can use the Rydberg formula:

$\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

where:
• $\lambda$ is the wavelength of the emitted spectral line,
• $R$ is the Rydberg constant,
• $Z$ is the atomic number (for a hydrogen atom, $Z=1$),
• $n_1$ is the principal quantum number of the lower energy level, and
• $n_2$ is the principal quantum number of the higher energy level.

For the Paschen series, the transitions end at the lower energy level $n_1=3$.

The energy of the emitted photon is inversely proportional to its wavelength:
$E=\frac{hc}{\lambda }$
Therefore, the longest wavelength (${\lambda }_{\text{max}}$) corresponds to the transition with the minimum energy change. The transition that yields the minimum energy change to the level $n_1=3$ is from the immediately adjacent level, $n_2=4$.

Substituting $Z=1$, $n_1=3$, and $n_2=4$ into the Rydberg formula, we get:

$\frac{1}{{\lambda }_{\text{max}}}=R[\frac{1}{3^2}-\frac{1}{4^2}]$

$\frac{1}{{\lambda }_{\text{max}}}=R[\frac{1}{9}-\frac{1}{16}]$

Find a common denominator for the fractions inside the brackets:
$\frac{1}{{\lambda }_{\text{max}}}=R\left[\frac{16-9}{144}\right]$

$\frac{1}{{\lambda }_{\text{max}}}=\frac{7R}{144}$

Taking the reciprocal to find the longest wavelength ${\lambda }_{\text{max}}$:
${\lambda }_{\text{max}}=\frac{144}{7R}$

Comparing this result with the given expression $\frac{144}{xR}$ (or $144/\left(xR\right)$, written in the question as 144R/x where R is in the denominator of the wavelength expression as shown by the formula):
$x=7$