Question

For sodium light, the two yellow lines occur at ${λ}_1$ and ${λ}_2$ wavelengths. If the mean of these two is $6000\stackrel{{}^{∘}}{A}$ and $âˆ\pounds {λ}_2−{λ}_1âˆ\pounds =6\stackrel{{}^{∘}}{A}$ , then the approximate energy difference between the two levels corresponding to ${λ}_1$ and ${λ}_2$ is

• A. $2×10^{−3}\text{e}\text{V}$
• B. $2\text{e}\text{V}$
• C. $2000\text{e}\text{V}$
• D. $2×10^{−6}\text{e}\text{V}$

Answer 1

Answer: (A)

Given, mean of ${λ}_1$ and ${λ}_2=6000Å$
$âˆ\pounds {λ}_2−{λ}_1âˆ\pounds =6â„«$
i.e. $\frac{{λ}_1+{λ}_2}{2}=6000Å$
${λ}_1+{λ}_2=12000Å$ ...(i)
${λ}_2−{λ}_1=6Å$ ...(ii)
Equating Equation (i) and (ii), we get
$2{λ}_1=12006$
${λ}_2=\frac{12006}{2}=6003Å$
and ${λ}_1=5997Å$
Now, the energy difference is
$ΔE=hc\left(\frac{1}{{λ}_1}−\frac{1}{{λ}_2}\right)$
$ΔE=hc\left(\frac{1}{{λ}_1}−\frac{1}{{λ}_2}\right)$
$ΔE=6.6×10^{−34}×3×10^8×\left(\frac{{λ}_2−{λ}_1}{{λ}_1{λ}_2}\right)$
$=\frac{6.6×10^{−34}×3×10^8×6×10^{−10}}{5997×10^{−10}×6003×10^{−10}×1.6×10^{−19}}\text{e}{\text{V}}^{}$
$=2.12×10^{−3}\text{e}\text{V}∼eq2×10^{−3}\text{e}\text{V}$