QJEEM24Ph90ca: JEE Main 2024 Physics - Class 12 JEEM24Ph90ca Solution: The longest wavelength associated with Paschen series is :…

To determine the longest wavelength associated with the Paschen series, we need to understand what the Paschen series is and how its wavelength can be calculated. The Paschen series pertains to the spectral line emissions of the hydrogen atom as an electron transitions from higher energy levels (n > 3) down to n = 3. The formula used to calculate the wavelength ( $\lambda$ ) of the emitted photon during such a transition in the hydrogen atom is given by the Rydberg formula:

$\frac{1}{\lambda} = R_{H} \left( \frac{1}{3^2} - \frac{1}{n^2} \right)$

where,

• $R_{H}$ is the Rydberg constant for hydrogen, $R_{H} = 1.097 \times 10^7 \, \text{m}^{-1}$ .
• $n$ is the principal quantum number of the higher energy level (For the longest wavelength, we use the smallest possible value of $n$ that is larger than 3, which is $n=4$ , since for longer wavelength transitions, the difference in energy levels should be the smallest).

To find the longest wavelength, we plug in $n=4$ into the Rydberg equation, as this corresponds to the smallest energy transition within the Paschen series ( $n=4$ to $n=3$ ):

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{16} \right)$

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{16 - 9}{144} \right)$

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144}$

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144}$

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144}$

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144} = 76403.47\, \text{m}^{-1}$

Finally, we calculate $\lambda$ by taking the reciprocal of this value:

$\lambda = \frac{1}{76403.47} \approx 1.308 \times 10^{-5} \, \text{m} = 1.308 \times 10^{-5} \times 10^{2} \, \text{cm} = 1.308 \times 10^{-3} \, \text{cm}$

It seems there was a mistake in my calculation. Revising the calculation properly:

$\lambda = \frac{1}{1.097 \times 10^7 \times \frac{7}{144}}$

Correctly evaluating this, we should directly compute:

$\lambda \approx \frac{144}{7 \times 1.097 \times 10^7} = \frac{144}{7.679 \times 10^7} \approx 1.876 \times 10^{-6} \, \text{m}$

Therefore, the correct option that matches our calculation for the longest wavelength in the Paschen series is:

Option B: $1.876 \times 10^{-6} \, \text{m}$ .