Question

The longest wavelength doublet absorption transition is observed at $589$ and $589.6\,nm$. Energy difference between two excited states is

• A. $3.31 \times 10^{-22}\,kJ$
• B. $3.31 \times 10^{-22}\,J$
• C. $2.98 \times 10^{-21}\,J$
• D. $3.0 \times 10^{-21}\,kJ$

Answer 1

Answer: (B)

Step I : Wavelength $(\lambda) = 589\,nm$
$= 589 \times 10^{-9}\,m$
Frequency $\left(\upsilon\right)=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{589 \times 10^{-9}}$
$=5.093\times10^{14}$ cycles per sec
Step II : Wavelength $\left(\lambda\right)=589.6\,nm=589.6\times10^{-9}\,m$
$\therefore \upsilon=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{589.6 \times} 10^{-9}$
$=5.088\times10^{14}$ cycles per sec
Energy difference between two excited states,
$\Delta E=6.626\times10^{-34} \left(5.093-5.088\right)10^{14}$
$\Delta x.m\Delta v=\frac{h}{4\pi}$
Uncertainty in the position,
$\Delta x=\frac{h}{4\pi m\Delta v}$
$=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 10 \times 10^{-3} \times 10}$
$=5.27\times10^{-34}\,m$