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Question: The Sodium yellow doublet has wavelengths of \(5890\overset{0}{\mathop{\text{A}}}\,\), \(\lambda \ov...

The Sodium yellow doublet has wavelengths of 5890A05890\overset{0}{\mathop{\text{A}}}\,, λA0\lambda \overset{0}{\mathop{\text{A}}}\,and resolving power of a grating to resolve these lines is 982, then value of λ\lambda is:
(A)5896A0 (B)5880A0 (C)5869A0 (D)5876A0 \begin{aligned} & (A)5896\overset{0}{\mathop{\text{A}}}\, \\\ & (B)5880\overset{0}{\mathop{\text{A}}}\, \\\ & (C)5869\overset{0}{\mathop{\text{A}}}\, \\\ & (D)5876\overset{0}{\mathop{\text{A}}}\, \\\ \end{aligned}

Explanation

Solution

In the above problem, we need to resolve the sodium doublets as separate wavelengths when one of them is known by grating. Now, we can use the definition of resolving power which is given by, wavelength upon the difference in wavelengths is equal to the resolving power needed to resolve two different lines. This shall be used to find the other wavelength of the Sodium Yellow Doublet.

Complete step-by-step answer:
Let us first assign some terms that we are going to use later in our calculation.
Let the resolving power of the grating be denoted by ‘N’. Then, its value is known to us as:
N=982\Rightarrow N=982
Now,
Let the two wavelengths of sodium yellow doublet be λ0{{\lambda }_{0}}and λ\lambda . Then, the value of λ0{{\lambda }_{0}}is given to us as:
λ0=5890A0\Rightarrow {{\lambda }_{0}}=5890\overset{0}{\mathop{\text{A}}}\,
And the value of λ\lambda is to be calculated. This can be done by writing the formula for resolving power of the grating. The resolving power of the grating can be expressed mathematically as follows:
N=λ0λλ0\Rightarrow N=\dfrac{{{\lambda }_{0}}}{\left| \lambda -{{\lambda }_{0}} \right|}
Putting the values of all the known terms in the above equation, we get:
982=5890λλ0 λ5890=5890982 λ5890=5.998 \begin{aligned} & \Rightarrow 982=\dfrac{5890}{\left| \lambda -{{\lambda }_{0}} \right|} \\\ & \Rightarrow \left| \lambda -5890 \right|=\dfrac{5890}{982} \\\ & \Rightarrow \left| \lambda -5890 \right|=5.998 \\\ \end{aligned}
Opening the modulus with positive sign, we get:
λ=5890+5.998 λ5896A0 \begin{aligned} & \Rightarrow \lambda =5890+5.998 \\\ & \therefore \lambda \approx 5896\overset{0}{\mathop{\text{A}}}\, \\\ \end{aligned}
And opening the modulus with negative sign, we get:
λ=58905.998 λ5884A0 \begin{aligned} & \Rightarrow \lambda =5890-5.998 \\\ & \therefore \lambda \approx 5884\overset{0}{\mathop{\text{A}}}\, \\\ \end{aligned}
Since, only one of them is mentioned in the options.
The final wavelength of the sodium yellow doublet will be equal to 5896A05896\overset{0}{\mathop{\text{A}}}\, .

So, the correct answer is “Option A”.

Note: In the last step of our solution, we calculated two different wavelengths. Since, there is no mention of first wavelength (λ0{{\lambda }_{0}}) being greater or less than the second wavelength (λ\lambda ), both of them can be the unknown wavelength of the sodium yellow doublet, but not simultaneously together.