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Chemistry Question on Bohr’s Model for Hydrogen Atom

The longest wavelength doublet absorption transition is observed at 589589 and 589.6 nm589.6\,nm. Energy difference between two excited states is

A

3.31×10−22 kJ3.31 \times 10^{-22}\,kJ

B

3.31×10−22 J3.31 \times 10^{-22}\,J

C

2.98×10−21 J2.98 \times 10^{-21}\,J

D

3.0×10−21 kJ3.0 \times 10^{-21}\,kJ

Answer

3.31×10−22 J3.31 \times 10^{-22}\,J

Explanation

Solution

Wavelength (λ)=589 nm(\lambda) = 589\,nm =589×10−9 m= 589 \times 10^{-9}\,m Frequency (υ)=cλ=3×108589×10−9\left(\upsilon\right)=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{589 \times 10^{-9}} =5.093×1014=5.093\times10^{14} cycles per sec Wavelength (λ)=589.6 nm=589.6×10−9 m\left(\lambda\right)=589.6\,nm=589.6\times10^{-9}\,m ∴υ=cλ=3×108589.6×10−9\therefore \upsilon=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{589.6 \times} 10^{-9} =5.088×1014=5.088\times10^{14} cycles per sec Energy difference between two excited states, ΔE=6.626×10−34(5.093−5.088)1014\Delta E=6.626\times10^{-34} \left(5.093-5.088\right)10^{14} Δx.mΔv=h4π\Delta x.m\Delta v=\frac{h}{4\pi} Uncertainty in the position, Δx=h4πmΔv\Delta x=\frac{h}{4\pi m\Delta v} =6.626×10−344×3.14×10×10−3×10=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 10 \times 10^{-3} \times 10} =5.27×10−34 m=5.27\times10^{-34}\,m