Question

What is the maximum wavelength of line of Balmer series of hydrogen spectrum?
(R = 1.09 × 10⁷ m^-1)

• A. 400 nm
• B. 660 nm
• C. 486 nm
• D. 434 nm

Answer 1

Answer: (B)

For maximum wavelength in the Balmer series, n₂ = 3 and n₁ = 2.
$\therefore \frac{1}{\lambda }=1.09\times 10^7\times 1^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$
$\Rightarrow \frac{1}{\lambda }=1.09\times 10^7\times 1\left(\frac{1}{4}-\frac{1}{9}\right)$
$=1.09\times 10^7\times 1\left(\frac{5}{36}\right)$
$\Rightarrow \lambda =1.09\times 10^7\times 1\left(\frac{5}{36}\right)$
$=6.60\times 10^{-7}m=660$ nm