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Chemistry Question on Structure of atom

Based on the equation : ΔE=−2.0×10−18J(1n22−1n12)\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 : (h=6.625×10−34 Js,C=3×108 ms−1)\left(h = 6.625 \times 10^{-34} \,Js, C= 3\times10^{8} \,ms^{-1}\right)

A

1.325×10−7m1.325 \times 10^{-7} m

B

1.325×10−10m1.325 \times 10^{-10} m

C

2.650×10−7m2.650 \times 10^{-7} m

D

5.300×10−10m5.300 \times 10^{-10} m

Answer

1.325×10−7m1.325 \times 10^{-7} m

Explanation

Solution

1λ=2×10−18hc[1(1)2−1(2)2]\frac{1}{\lambda}=\frac{2\times10^{-18}}{h c} \left[\frac{1}{\left(1\right)^{2}}-\frac{1}{\left(2\right)^{2}}\right]
⇒1λ=2×10−186.625×10−34×3×108×34\Rightarrow\quad\frac{1}{\lambda}=\frac{2\times10^{-18}}{6.625\times10^{-34}\times3\times10^{8}} \times\frac{3}{4}
⇒λ=2×6.625×10−34×10810−18\Rightarrow\quad\lambda=\frac{2\times6.625\times10^{-34}\times10^{8}}{10^{-18}}
=13.25×10−8=13.25\times10^{-8}
=13.25×10−7 m=13.25\times10^{-7} \,m