The limiting line in Balmer series will have a frequency of

Question

The limiting line in Balmer series will have a frequency of (A) 32.29 × 1015 s -1 (B) 3.65 × 1014 s -1 (C) -8.22 × 1014 s -1 (D) 8.22 × 1014 s -1

Tardigrade Admin · Accepted Answer

The limiting line of Balmer series refers to the transition from ∞ to 2 nd orbit. v =3.29 × 1015((1/n12)-(1/n22)) s-1(n1=2, n=∞) =3.29 × 1015 × (1/4)=8.22 × 1014 s -1

Q. The limiting line in Balmer series will have a frequency of

$32.29 \times 1 0^{15} s^{- 1}$

$3.65 \times 1 0^{14} s^{- 1}$

$- 8.22 \times 1 0^{14} s^{- 1}$

$8.22 \times 1 0^{14} s^{- 1}$

The limiting line of Balmer series refers to the transition from $\infty$ to 2 nd orbit.
$v = 3.29 \times 1 0^{15} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) s^{- 1} (n_{1} = 2, n = \infty)$
$= 3.29 \times 1 0^{15} \times \frac{1}{4} = 8.22 \times 1 0^{14} s^{- 1}$

Q. The limiting line in Balmer series will have a frequency of

32.29×1015s−1

3.65×1014s−1

−8.22×1014s−1

8.22×1014s−1

The limiting line of Balmer series refers to the transition from ∞ to 2 nd orbit. v=3.29×1015(n12​1​−n22​1​)s−1(n1​=2,n=∞) =3.29×1015×41​=8.22×1014s−1

$32.29 \times 1 0^{15} s^{- 1}$

$3.65 \times 1 0^{14} s^{- 1}$

$- 8.22 \times 1 0^{14} s^{- 1}$

$8.22 \times 1 0^{14} s^{- 1}$

The limiting line of Balmer series refers to the transition from $\infty$ to 2 nd orbit.
$v = 3.29 \times 1 0^{15} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) s^{- 1} (n_{1} = 2, n = \infty)$
$= 3.29 \times 1 0^{15} \times \frac{1}{4} = 8.22 \times 1 0^{14} s^{- 1}$