Question

The Frequency Of One Of The Lines In Paschen Series Of A Hydrogen Atom Is 2.34 * 10(14) Hz. The Quantum Number, no Wch Produces Ts Transition Is i) 3 ii) 4 iii) 6 iv) 5

Answer 1

The answer of the question that you asked is Option (iv)

Answer 2

Answer: The correct answer is (iv).

Explanation:

Energy associated to the $n^{th}$ - energy level will be:

$E_n=\frac{E_o\times Z^2}{n^2}$      ...(1)

$E_o=-2.18\times 10^{-18}J$

Paschen series have n = 3

For the transition from n = 3 to n = n, we use equation 1 for the difference in energy.

Using Planck's equation,

$E=h\nu$

where, $h=6.624\times 10^{-34}Js$   (Planck's constant)

$\nu=Frequency$

Difference in energy is written as:

$\Delta E=E_n-E_3$

Putting values in above equation, we get

$(6.624\times 10^{-34}Js)(2.34\times 10^{14}sec^{-1})=\left(\frac{(-2.18\times 10^{-18})J\times 1^2}{n^2}\right)-\left(\frac{(-2.18\times 10^{-18}J)\times 1^2}{3^2}\right)$

$n^2=25$

$n=\pm 5$

Negative sign is cancelled, because energy level can't be on the negative side.

Hence, the quantum number which represents this transition is 5.