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Chemistry Question on Bohr’s Model for Hydrogen Atom

The energy of photon is given as : Δ\Delta E/atom= 3.03×10−19Jatom−1,3.03 \times 10^{-19} J atom^{-1}, then the wavelength (λ)(\lambda) of the photon is (Given, h(Planck's constant) =6.63×10−34=6.63 \times 10^{-34} Js, c (velocity of light) =3.00×108ms−1= 3.00 \times 10^8 ms^{-1}

A

6.56 nm

B

65.7 nm

C

656 nm

D

0.656 nm

Answer

656 nm

Explanation

Solution

According to formula,E=hcλ(v=cλ),E=\frac{hc}{\lambda}\Bigg(v=\frac{c}{\lambda}\Bigg) \hspace10mm Energy E = hv \hspace10mm 3.03 \times 10^{-19}=\frac{hc}{\lambda}               λ=6.63×10−34×3.0×1083.03×10−19\, \, \, \, \, \, \, \, \, \, \, \, \, \, \lambda=\frac{6.63\times10^{-34}\times3.0\times10^8}{3.03\times10^{-19}} =6.56×10−7m= 6.56 \times 10^{-7} m =6.56×10−7×109nm= 6.56 \times 10^{-7} \times 10^9 nm =6.56×102nm=656nm= 6.56 \times 10^2 nm = 656 nm