Question

Choose the correct statement (s) for hydrogen and deuterium atoms considering motion of reaction

• A. The radius of first Bohr orbit of deuterium is less than that of hydrogen
• B. The speed of electron in the first Bohr orbit of deuterium is more that that of hydrogen
• C. The wave length of first Balmer line of deuterium is more than that of hydrogen
• D. The angular momentum of electron in the first Bohr orbit of deuterium is more than that of hydrogen

Answer 1

The correct option is D The angular momentum of electron in the first Bohr orbit of deuterium is more than that of hydrogen

$\begin{array}{c}\text{Bohr's considerd the nucleus is at rest and only electron is revolving around it.}\\ \text{but here in question motion of nucleus is given.}\end{array}$

$\begin{array}{c}\text{Hydrogen}{=}_1{H}^2\\ \text{deuterium}{=}_1{H}^2\end{array}$

$\begin{array}{c}\text{Now, if we consider motion of nucleus}\\ \text{also, then we have to consider nucleus mass}\\ \text{also.}\end{array}$

$\text{(a)}\text{Now,}r\propto \frac{n^2}{m}$

$\begin{array}{c}\text{So, radius of deuterium is less than that of hydrogen, option (a) is correct}\end{array}$

$\text{(b) velocity}\propto \frac{z^2}{n^{2}m}$

$\begin{array}{c}\text{So, velocity of electron in hydrogen is more, option B incorrect}\end{array}$

$\begin{array}{c}\text{(c) wavelength}\\ \frac{1}{\lambda }=R{z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\\ \text{and}\therefore R\propto m\end{array}$

$\begin{array}{cc}& \frac{1}{\lambda }\propto R\\ & =\frac{1}{\lambda }\propto m\end{array}$

$\begin{array}{c}\text{So, wavelength of hydrogen will be more than that of deuterium,}\end{array}$

$\text{option c is incorrect}$

$\begin{array}{c}\text{(d) as angular momentum is conserved always inspite of mass and atomic no. , it only depend on Energy level.}\\ \text{mvr}=\frac{nh}{2\pi }\end{array}$

$\text{So, option}\left(d\right)\text{is incorrect.}$.