
Some of the wavelengths observed in the emission spectrum of neutral hydrogen gas are: 912,1026, 1216, 3646, 6563 $A^o$. If broadband light is passing through neutral hydrogen gas at room temperature, the wavelength that will not be absorbed strongly is?
A. 1026 $A^o$
B. 1216 $A^o$
C. 912 $A^o$
D. 3646 $A^o$
Answer
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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:
$\dfrac{1}{\lambda} = R_h(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} )$
We need to get the smallest and longest wavelength of the spectrum. So, what are the values of $n_1$ and $n_2$ for the smallest and longest wavelength? The highest wavelength will be for $n_1$ = 1 and $n_2$ = 2. The smallest wavelength will be for $n_1$ = 1 and $n_2$ = infinite. So let’s find the largest wavelength.
$\dfrac{1}{\lambda} = R_h(\dfrac{1}{1} - \dfrac{1}{4} )$
$1/\lambda = R_h(3/4)$
$\dfrac{1}{\lambda} = R_h(\dfrac{3}{4} )$
$\lambda = \dfrac{4}{3 R_h}$
The value of $R_h$ is
$R_h$ =$1.0967 \times 10^7 m^{-1}$
$\lambda = \dfrac{4}{3 \times 1.0967 \times 10^7}$
$\lambda = 1.2158 \times 10^{-7}$
$\lambda = 12158 \times 10^{-10}$
$\lambda = 1216 A^o$
Similarly, if we calculate the shortest wavelength
$\dfrac{1}{\lambda} = R_h(\dfrac{1}{1} )$
$\lambda = \dfrac{1}{R_h}$
The value of $R_h$ is
$R_h$ = $1.0967 \times 10^7 m^{-1}$
$\lambda = \dfrac{1}{1.0967 \times 10^{7}}$
$\lambda = 0.9118 \times 10^{-7}$
$\lambda = 912 A^o$
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 $\dfrac{1}{\lambda}$ value as it is often seen in calculations. The value of the $\dfrac{1}{\lambda}$ value is 912A. We can also find the frequency range from there we can find the wavelength range.
Complete step-by-step answer:
Rydberg formula for Hydrogen is:
$\dfrac{1}{\lambda} = R_h(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} )$
We need to get the smallest and longest wavelength of the spectrum. So, what are the values of $n_1$ and $n_2$ for the smallest and longest wavelength? The highest wavelength will be for $n_1$ = 1 and $n_2$ = 2. The smallest wavelength will be for $n_1$ = 1 and $n_2$ = infinite. So let’s find the largest wavelength.
$\dfrac{1}{\lambda} = R_h(\dfrac{1}{1} - \dfrac{1}{4} )$
$1/\lambda = R_h(3/4)$
$\dfrac{1}{\lambda} = R_h(\dfrac{3}{4} )$
$\lambda = \dfrac{4}{3 R_h}$
The value of $R_h$ is
$R_h$ =$1.0967 \times 10^7 m^{-1}$
$\lambda = \dfrac{4}{3 \times 1.0967 \times 10^7}$
$\lambda = 1.2158 \times 10^{-7}$
$\lambda = 12158 \times 10^{-10}$
$\lambda = 1216 A^o$
Similarly, if we calculate the shortest wavelength
$\dfrac{1}{\lambda} = R_h(\dfrac{1}{1} )$
$\lambda = \dfrac{1}{R_h}$
The value of $R_h$ is
$R_h$ = $1.0967 \times 10^7 m^{-1}$
$\lambda = \dfrac{1}{1.0967 \times 10^{7}}$
$\lambda = 0.9118 \times 10^{-7}$
$\lambda = 912 A^o$
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 $\dfrac{1}{\lambda}$ value as it is often seen in calculations. The value of the $\dfrac{1}{\lambda}$ value is 912A. We can also find the frequency range from there we can find the wavelength range.
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