Question

Based on the equation :
$\Delta E = -2.0 \times 10^{-18} J \left(\frac{1}{n^{2}_{2}}-\frac{1}{n^{2}_{1}}\right)$
the wavelength of the light that must be absorbed to excite hydrogen electron from level n=l to level n=2 will be :
$\left(h = 6.625 \times 10^{-34} \,Js, C= 3\times10^{8} \,ms^{-1}\right)$

• 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)

$\frac{1}{\lambda}=\frac{2\times10^{-18}}{h c} \left[\frac{1}{\left(1\right)^{2}}-\frac{1}{\left(2\right)^{2}}\right]$

$\Rightarrow \quad\frac{1}{\lambda}=\frac{2\times10^{-18}}{6.625\times10^{-34}\times3\times10^{8}} \times\frac{3}{4}$

$\Rightarrow \quad\lambda=\frac{2\times6.625\times10^{-34}\times10^{8}}{10^{-18}}$

$=13.25\times10^{-8}$

$=13.25\times10^{-7} \,m$