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Question: Some of the wavelengths observed in the emission spectrum of neutral hydrogen gas are: 912,1026, 121...

Some of the wavelengths observed in the emission spectrum of neutral hydrogen gas are: 912,1026, 1216, 3646, 6563 AoA^o. If broadband light is passing through neutral hydrogen gas at room temperature, the wavelength that will not be absorbed strongly is?
A. 1026 AoA^o
B. 1216 AoA^o
C. 912 AoA^o
D. 3646 AoA^o

Explanation

Solution

Hint: We need to find the range of wavelengths that we will be absorbed in the hydrogen spectrum. For this, we will apply the Rydberg formula and obtain the range. Once we get the range it will be easy to check the wrong option i.e the wavelength that will not be absorbed strongly.

Complete step-by-step answer:
Rydberg formula for Hydrogen is:
1λ=Rh(1n121n22)\dfrac{1}{\lambda} = R_h(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} )
We need to get the smallest and longest wavelength of the spectrum. So, what are the values of n1n_1 and n2n_2 for the smallest and longest wavelength? The highest wavelength will be for n1n_1 = 1 and n2n_2 = 2. The smallest wavelength will be for n1n_1 = 1 and n2n_2 = infinite. So let’s find the largest wavelength.
1λ=Rh(1114)\dfrac{1}{\lambda} = R_h(\dfrac{1}{1} - \dfrac{1}{4} )
1/λ=Rh(3/4)1/\lambda = R_h(3/4)
1λ=Rh(34)\dfrac{1}{\lambda} = R_h(\dfrac{3}{4} )
λ=43Rh\lambda = \dfrac{4}{3 R_h}
The value of RhR_h is
RhR_h =1.0967×107m11.0967 \times 10^7 m^{-1}
λ=43×1.0967×107\lambda = \dfrac{4}{3 \times 1.0967 \times 10^7}
λ=1.2158×107\lambda = 1.2158 \times 10^{-7}
λ=12158×1010\lambda = 12158 \times 10^{-10}
λ=1216Ao\lambda = 1216 A^o
Similarly, if we calculate the shortest wavelength
1λ=Rh(11)\dfrac{1}{\lambda} = R_h(\dfrac{1}{1} )
λ=1Rh\lambda = \dfrac{1}{R_h}
The value of RhR_h is
RhR_h = 1.0967×107m11.0967 \times 10^7 m^{-1}
λ=11.0967×107\lambda = \dfrac{1}{1.0967 \times 10^{7}}
λ=0.9118×107\lambda = 0.9118 \times 10^{-7}
λ=912Ao\lambda = 912 A^o
The range of the wavelengths is 911A - 1216A
In the given options 3646A is not in the range. So it will not be absorbed strongly.

Note: It is very handy to remember 1λ\dfrac{1}{\lambda} value as it is often seen in calculations. The value of the 1λ\dfrac{1}{\lambda} value is 912A. We can also find the frequency range from there we can find the wavelength range.