Given the value of Rydberg constant is 107 m-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be :

Question

Given the value of Rydberg constant is 107 m-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be : (A) 0.5 × 107 m-1 (B) 0.25 × 107 m-1 (C) 2.5 × 107 m-1 (D) 0.025 × 104 m-1

Tardigrade Admin · Accepted Answer

(1/λ ) = RZ2 ((1/n22) - (1/n12)) = 107 ×12 ((1/22) - (1/ ∞2)) ⇒ wave number = (1/λ ) = 0.25 ×107 m-1

Q. Given the value of Rydberg constant is $1 0^{7} m^{- 1}$ , the wave number of the last line of the Balmer series in hydrogen spectrum will be :

$0.5 \times 1 0^{7} m^{- 1}$

$0.25 \times 1 0^{7} m^{- 1}$

$2.5 \times 1 0^{7} m^{- 1}$

$0.025 \times 1 0^{4} m^{- 1}$

$\frac{1}{λ} = R Z^{2} (\frac{1}{n _{2}^{2}} - \frac{1}{n _{1}^{2}}) = 1 0^{7} \times 1^{2} (\frac{1}{2 ^{2}} - \frac{1}{\infty ^{2}})$
$\Rightarrow$ wave number $= \frac{1}{λ} = 0.25 \times 1 0^{7} m^{- 1}$

Q. Given the value of Rydberg constant is 107m−1, the wave number of the last line of the Balmer series in hydrogen spectrum will be :

0.5×107m−1

0.25×107m−1

2.5×107m−1

0.025×104m−1