Question

Some of the wavelengths observed in the emission spectrum of neutral hydrogen gas are: $912,1026,1216,3646,6563\stackrel{o}{A}$. If broad band light is passing through neutral hydrogen gas at room temperature, the wavelength that will not be absorbed strongly is?

• A. $1026\stackrel{o}{A}$
• B. $1216\stackrel{o}{A}$
• C. $912\stackrel{o}{A}$
• D. $3646\stackrel{o}{A}$

Answer 1

Answer: (D)

$3646\stackrel{o}{A}$

$\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Here $n_1=1and{n}_2=2$

$\frac{1}{\lambda }=R_H(1-\frac{1}{4})$

$\lambda =\frac{4}{3R_H}$ is largest wavelength

Now taking $n_1=1and{n}_2=\infty$

$\frac{1}{\lambda }=R_H(1-0)$

$\lambda =\frac{1}{R_H}$ is smallest wavelength

Now putting value of $R=1.09\times {10}^{-2}$

We will get the range as 915 Armstrong to 1223 Armstrong.

So out of given option 3646 A is out of this range so D option is correct answer.