The first line of Balmer series has wavelength 6563 mathringA. What will be the wavelength of the first member of Lyman series

Question

The first line of Balmer series has wavelength 6563 mathringA. What will be the wavelength of the first member of Lyman series (A) 1215.4 mathringA (B) 2500 mathringA (C) 7500 mathringA (D) 600 mathringA

Tardigrade Admin · Accepted Answer

(1/Balmer) R (1/22) (1/32) =(5R/36), (1/Lyman) R (1/12) (1/22) =(3R/4) ∴ λLyman = λBalmer×(5/27)= 1215.4 mathringA

Q. The first line of Balmer series has wavelength $6563 \overset{˚}{A}$ . What will be the wavelength of the first member of Lyman series

$1215.4 \overset{˚}{A}$

$2500 \overset{˚}{A}$

$7500 \overset{˚}{A}$

$600 \overset{˚}{A}$

$\frac{1}{B a l m er} R \frac{1}{2 ^{2}} \frac{1}{3 ^{2}} = \frac{5 R}{36},$

$\frac{1}{L y man} R \frac{1}{1 ^{2}} \frac{1}{2 ^{2}} = \frac{3 R}{4}$

$∴ λ_{L y man} = λ_{B a l m er} \times \frac{5}{27} = 1215.4 \overset{˚}{A}$

Q. The first line of Balmer series has wavelength 6563A˚. What will be the wavelength of the first member of Lyman series

1215.4A˚

2500A˚

7500A˚

600A˚