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Physics Question on Atoms

For sodium light, the two yellow lines occur at λ1\lambda _{1} and λ2\lambda _{2} wavelengths. If the mean of these two is 6000 A∘6000 \, \overset{^\circ }{A} and ∣λ2−λ1∣=6 A∘\left|\lambda _{2} - \lambda _{1}\right|=6 \, \overset{^\circ }{A} , then the approximate energy difference between the two levels corresponding to λ1\lambda _{1} and λ2\lambda _{2} is

A

2×10−3 eV2\times 10^{- 3} \, \text{e}\text{V}

B

2 eV2 \, \text{e}\text{V}

C

2000 eV2000 \, \text{e}\text{V}

D

2×10−6 eV2\times 10^{- 6} \, \text{e}\text{V}

Answer

2×10−3 eV2\times 10^{- 3} \, \text{e}\text{V}

Explanation

Solution

Given, mean of λ1\lambda _{1} and λ2=6000 ?\lambda _{2}=6000 \, ? ∣λ2−λ1∣=6 ?\left|\lambda _{2} - \lambda _{1}\right|=6 \, ? i.e. λ1+λ22=6000 ?\frac{\lambda _{1} + \lambda _{2}}{2}=6000 \, ? λ1+λ2=12000 ?\lambda _{1}+\lambda _{2}=12000 \, ? ...(i) λ2−λ1=6 ?\lambda _{2}-\lambda _{1}=6 \, ? ...(ii) Equating Equation (i) and (ii), we get 2λ1=120062\lambda _{1}=12006 λ2=120062=6003 ?\lambda _{2}=\frac{12006}{2}=6003 \, ? and λ1=5997 ?\lambda _{1}=5997 \, ? Now, the energy difference is ΔE=hcλ1−hcλ2\Delta E=\frac{h c}{\lambda _{1}}-\frac{h c}{\lambda _{2}} ΔE=hc(1(λ)1−1(λ)2)\Delta E=hc\left(\frac{1}{\left(\lambda \right)_{1}} - \frac{1}{\left(\lambda \right)_{2}}\right) ΔE=6.6×(10)−34×3×(10)8×((λ)2−(λ)1(λ)1(λ)2)\Delta E=6.6\times \left(10\right)^{- 34}\times 3\times \left(10\right)^{8}\times \left(\frac{\left(\lambda \right)_{2} - \left(\lambda \right)_{1}}{\left(\lambda \right)_{1} \left(\lambda \right)_{2}}\right) =6.6×10−34×3×108×6×10−105997×10−10×6003×10−10×1.6×10−19 eV =\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}^{ \, } =2.12×10−3 eV∼eq2×10−3 eV=2.12\times 10^{- 3} \, \text{e}\text{V}\sim eq2\times 10^{- 3} \, \text{e}\text{V}