Question

For sodium light, the two yellow lines occur at $\lambda _{1}$ and $\lambda _{2}$ wavelengths. If the mean of these two is $6000 \, \overset{^\circ }{A}$ and $\left|\lambda _{2} - \lambda _{1}\right|=6 \, \overset{^\circ }{A}$ , then the approximate energy difference between the two levels corresponding to $\lambda _{1}$ and $\lambda _{2}$ is

• A. $2\times 10^{- 3} \, \text{e}\text{V}$
• B. $2 \, \text{e}\text{V}$
• C. $2000 \, \text{e}\text{V}$
• D. $2\times 10^{- 6} \, \text{e}\text{V}$

Answer 1

Answer: (A)

Given, mean of $\lambda _{1}$ and $\lambda _{2}=6000 \, Å$
$\left|\lambda _{2} - \lambda _{1}\right|=6 \, Å$
i.e. $\frac{\lambda _{1} + \lambda _{2}}{2}=6000 \, Å$
$\lambda _{1}+\lambda _{2}=12000 \, Å$ ...(i)
$\lambda _{2}-\lambda _{1}=6 \, Å$ ...(ii)
Equating Equation (i) and (ii), we get
$2\lambda _{1}=12006$
$\lambda _{2}=\frac{12006}{2}=6003 \, Å$
and $\lambda _{1}=5997 \, Å$
Now, the energy difference is
$\Delta E=h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)$
$\Delta E=h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)$
$\Delta E=6.6 \times 10^{-34} \times 3 \times 10^8 \times\left(\frac{\lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right)$
$=\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8} \times 6 \times 10^{- 10}}{5997 \times 10^{- 10} \times 6003 \times 10^{- 10} \times 1.6 \times 10^{- 19}} \, \text{e} \text{V}^{ \, }$
$=2.12\times 10^{- 3} \, \text{e}\text{V}\sim eq2\times 10^{- 3} \, \text{e}\text{V}$