The longest wavelength associated with Paschen series is

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Q.
The longest wavelength associated with Paschen series is

$(Given R_{H} = 1.097 \times 10^{7} in SI unit)$ [2024]

1 $2.973 \times 10^{- 6} m$

2 $3.646 \times 10^{- 6} m$

3 $1.094 \times 10^{- 6} m$

4 $1.876 \times 10^{- 6} m$

Ans.
(4)

For longest wavelength in Paschen's series:

$\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

   $n_{1} = 3 and n_{2} = 4$

   $\frac{1}{λ} = R [\frac{1}{{(3)}^{2}} - \frac{1}{{(4)}^{2}}] = R [\frac{16 - 9}{16 \times 9}]$

   $\Rightarrow λ = \frac{16 \times 9}{7 R} = \frac{16 \times 9}{7 \times 1.097 \times 10^{7}} = 1.876 \times 10^{- 6} m$

Solution

Q.
The longest wavelength associated with Paschen series is

$(Given R_{H} = 1.097 \times 10^{7} in SI unit)$ [2024]

1 $2.973 \times 10^{- 6} m$

2 $3.646 \times 10^{- 6} m$

3 $1.094 \times 10^{- 6} m$

4 $1.876 \times 10^{- 6} m$

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Solution

Q. The longest wavelength associated with Paschen series is (Given RH=1.097×107 in SI unit) [2024]

1 2.973×10-6m

2 3.646×10-6m

3 1.094×10-6m

4 1.876×10-6m

Ans. (4) For longest wavelength in Paschen's series: 1λ=R[1n12-1n22] n1=3 and n2=4 1λ=R[1(3)2-1(4)2]=R[16-916×9] ⇒λ=16×97R=16×97×1.097×107=1.876×10-6m

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Q.
The longest wavelength associated with Paschen series is

$(Given R_{H} = 1.097 \times 10^{7} in SI unit)$ [2024]

1 $2.973 \times 10^{- 6} m$

2 $3.646 \times 10^{- 6} m$

3 $1.094 \times 10^{- 6} m$

4 $1.876 \times 10^{- 6} m$