Question

The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (given ionization energy of $H=2.18\times {10}^{-18}ato{m}^{-1}$ and $h=6.606\times {10}^{-34}j-s$)

• A. 1.54 X 10¹⁵ s^-1
• B. 1.03 X 10¹⁵ s^-1
• C. 3.08 X 10¹⁵ s^-1
• D. 2 X 10¹⁵ s^-1

Answer 1

The correct option is C 3.08 X 10¹⁵ s^-1

$v=C{R}_H{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$=3\times {10}^8\times 1.1\times {10}^7\times 1^2(\frac{1}{1^2}-\frac{1}{4^2})$
$=3.3\times {10}^{15}\left(\frac{15}{16}\right)$
$=3.3\times {10}^{15}\times 0.9375$
$=3.08\times {10}^{15}{s}^{-1}$