Question

Based on the equation :
$\Delta E=-2.0\times 10^{-18}J\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$
the wavelength of the light that must be absorbed to excite hydrogen electron from level n=l to level n=2 will be :
$\left(h=6.625\times 10^{-34}Js,C=3\times 10^{8}m{s}^{-1}\right)$

• A. $1.325\times 10^{-7}m$
• B. $1.325\times 10^{-10}m$
• C. $2.650\times 10^{-7}m$
• D. $5.300\times 10^{-10}m$

Answer 1

Answer: (A)

$\frac{1}{\lambda }=\frac{2\times 10^{-18}}{hc}\left[\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right]$

$\Rightarrow \frac{1}{\lambda }=\frac{2\times 10^{-18}}{6.625\times 10^{-34}\times 3\times 10^8}\times \frac{3}{4}$

$\Rightarrow \lambda =\frac{2\times 6.625\times 10^{-34}\times 10^8}{10^{-18}}$

$=13.25\times 10^{-8}$

$=13.25\times 10^{-7}m$