If wavelength corresponding to 2nd line of Lyman series is λ, then the wavelength corresponding to last line of Balmer series will be:

Question

If wavelength corresponding to 2nd line of Lyman series is λ, then the wavelength corresponding to last line of Balmer series will be: (A) (9 λ/8) (B) (32 λ/9) (C) (27 λ/16) (D) (24 λ/7)

Tardigrade Admin · Accepted Answer

(1/λ)=R Z2((1/12)-(1/32)) =(8 R Z2/9) (1/λ prime)=R Z2((1/22)-(1/(∞)2)) =(R Z2/4) (1/λ')=(9/32 λ) λ'=(32 λ/9)

Q. If wavelength corresponding to $2^{n d}$ line of Lyman series is $λ$ , then the wavelength corresponding to last line of Balmer series will be:

$\frac{9 λ}{8}$

$\frac{32 λ}{9}$

$\frac{27 λ}{16}$

$\frac{24 λ}{7}$

$\frac{1}{λ} = R Z^{2} (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}})$
$= \frac{8 R Z ^{2}}{9}$
$\frac{1}{λ ^{'}} = R Z^{2} (\frac{1}{2 ^{2}} - \frac{1}{( \infty ) ^{2}})$
$= \frac{R Z ^{2}}{4}$
$\frac{1}{λ ^{'}} = \frac{9}{32 λ}$
$λ^{'} = \frac{32 λ}{9}$

Q. If wavelength corresponding to 2nd line of Lyman series is λ, then the wavelength corresponding to last line of Balmer series will be:

89λ​

932λ​

1627λ​

724λ​