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Question: Na atom remains in lowest excited state energy for a time \(1.6\, \times \,{10^{ - 8}}\,s\) before i...

Na atom remains in lowest excited state energy for a time 1.6 × 10−8 s1.6\, \times \,{10^{ - 8}}\,s before it makes a transition to a ground state emitting a photon of wavelength 589 nm589\,nm . The wavelength spread corresponding to this line will be:
(A) 10−4 nm{10^{ - 4}}\,nm
(B) 10−5 nm{10^{ - 5}}\,nm
(C) 10−6 nm{10^{ - 6}}\,nm
(D) 10−2 nm{10^{ - 2}}\,nm

Explanation

Solution

Hint
You can easily solve the question using the Heisenberg’s Uncertainty Principle for energy and time, i.e. ΔE × Δt = ℏ2\Delta E\, \times \,\Delta t\, = \,\dfrac{\hbar }{2} , where ℏ = h2π\hbar \, = \,\dfrac{h}{{2\pi }} . Just convert the ΔE\Delta E in terms of λ\lambda before using the Uncertainty equation.

Complete step by step answer
As told in the hint, we are going to use Heisenberg’s Uncertainty Principle for energy and time to solve the question. But first, let’s have a look at what is given to us in the question.
Question has told us that uncertainty in time (Δt)\left( {\Delta t} \right) is = 1.6 × 10−8s = \,1.6\, \times \,{10^{ - 8}}s
The question has also told us that the wavelength (λ)\left( \lambda \right) is = 589nm589{\kern 1pt} nm
Equations that we will be using are:
Heisenberg’s Uncertainty Equation for energy and time:
ΔE × Δt = ℏ2\Delta E\, \times \,\Delta t\, = \,\dfrac{\hbar }{2}
And, the relation of energy with wavelength:
E = hcλE\, = \,\dfrac{{hc}}{\lambda } … (1)
But first we need to convert the Energy-Wavelength relation, we will do this by taking delta on both sides of the equation:
Δ(E) = Δ(hcλ)\Delta \left( E \right)\, = \,\Delta \left( {\dfrac{{hc}}{\lambda }} \right) … (2)
Using the basic differential rule, we resolve the equation to be:
ΔE = hcΔλλ2\Delta E\, = \,hc\dfrac{{\Delta \lambda }}{{{\lambda ^2}}} … (3)
Now that we have manipulated the Energy-Wavelength relation, we can proceed with the solution to our question.
Let’s simplify the Heisenberg’s Uncertainty Equation:
We know that ℏ= h2π\hbar {\kern 1pt} = \,\dfrac{h}{{2\pi }}
Now the Uncertainty equation becomes:
ΔE × Δt = (h2π)2\Delta E\, \times \,\Delta t\, = \,\dfrac{{\left( {\dfrac{h}{{2\pi }}} \right)}}{2}
ΔE × Δt = h4π\Delta E\, \times \,\Delta t\, = \,\dfrac{h}{{4\pi }}
Substituting the value of ΔE\Delta E from equation (3), that we deduced:
hc×Δλλ2 × Δt = h4πhc \times \dfrac{{\Delta \lambda }}{{{\lambda ^2}}}\, \times \,\Delta t\, = \,\dfrac{h}{{4\pi }}
We have to find the value of the wavelength spread, that is Δλ\Delta \lambda
After transposing, we get:
Δλ = λ24π × c × Δt\Delta \lambda \, = \,\dfrac{{{\lambda ^2}}}{{4\pi \, \times \,c\, \times \,\Delta t}}
Substituting the values of λ\lambda and Δt\Delta t that are given in the question, and the value of c (speed of light = 3 × 108m/s3\, \times \,{10^8}{\kern 1pt} m/s ), we get:
Δλ = (589 × 10−9)24π × 3 × 108 × 1.6 ×10−8\Delta \lambda \, = \,\dfrac{{{{\left( {589\, \times \,{{10}^{ - 9}}} \right)}^2}}}{{4\pi \, \times \,3\, \times \,{{10}^8}\, \times \,1.6\, \times {{10}^{ - 8}}}}
Solving this equation, we get:
Δλ = 10−4 nm\Delta \lambda \, = \,{10^{ - 4}}\,nm
Hence, option (A) is the correct answer.

Note
Many students confuse between whether the Uncertainty equation for energy and time has ℏ or h\hbar \,or\,h and thus do an error in the solution. Many times, students get stuck at the manipulation of the Energy-Wavelength Relation, so remember this manipulation as it is also widely used.