Question 127: Atoms & Nuclei

Question

Two electrons are moving in orbits of two hydrogen like atoms with speeds $3 	imes 10^5 \mathrm{~m} / \mathrm{s}$ and $2.5 	imes 10^5 \mathrm{~m} / \mathrm{s}$ respectively. If the radii of these orbits are nearly same then the possible order of energy states are $\_\_\_\_$ respectively.

Accepted Answer

The velocity of an electron in the $\mathrm{n}^{	ext {th }}$ orbit is proportional to the atomic number Z and inversely proportional to the principal quantum number n .
$$ \mathrm{v} \propto \frac{\mathrm{Z}}{\mathrm{n}} $$
$$\Rightarrow $$ $$ \mathrm{v}=\mathrm{k}_{\mathrm{v}} \cdot \frac{\mathrm{Z}}{\mathrm{n}} \ldots (i)$$
Where $\mathrm{k}_{\mathrm{v}}$ is constant for the velocity.
The radius of the $\mathrm{n}^{	ext {th }}$ orbit is proportional to the square of the principal quantum number n and inversely proportional to the atomic number Z.
$$ r \propto \frac{n^2}{Z} $$
$$\Rightarrow $$ $$ \mathrm{r}=\mathrm{k}_{\mathrm{r}} \cdot \frac{\mathrm{n}^2}{\mathrm{Z}} \ldots(\mathrm{ii}) $$
Let $\mathrm{r}_1$ be the radius for the first atom (with state $\mathrm{n}_1, \mathrm{Z}_1$ ) and $\mathrm{r}_2$ for the second (with state $\mathrm{n}_2, \mathrm{Z}_2$ ).
$$ \mathrm{r}_1=\mathrm{r}_2 $$
$$\Rightarrow $$ $\mathrm{k}_{\mathrm{r}} \cdot \frac{\mathrm{n}_1^2}{\mathrm{Z}_1}=\mathrm{k}_{\mathrm{r}} \cdot \frac{\mathrm{n}_2^2}{\mathrm{Z}_2}$
$$\Rightarrow $$ $\frac{\mathrm{n}_1^2}{\mathrm{Z}_1}=\frac{\mathrm{n}_2^2}{\mathrm{Z}_2}$
$$ \frac{\mathrm{Z}_1}{\mathrm{Z}_2}=\frac{\mathrm{n}_1^2}{\mathrm{n}_2^2}=\left(\frac{\mathrm{n}_1}{\mathrm{n}_2}ight)^2 \ldots (iii)$$
Let $\mathrm{v}_1$ and $\mathrm{v}_2$ be the speeds.
$$ \mathrm{v}_1=\mathrm{k}_{\mathrm{v}} \frac{\mathrm{Z}_1}{\mathrm{n}_1} \ldots (iv)$$
$\mathrm{v}_2=\mathrm{k}_{\mathrm{v}} \frac{\mathrm{Z}_2}{\mathrm{n}_2} \ldots(\mathrm{v})$
$$	herefore $$ $\frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{k}_{\mathrm{v}}\left(\frac{\mathrm{Z}_1}{\mathrm{n}_1}ight)}{\mathrm{k}_{\mathrm{v}}\left(\frac{\mathrm{Z}_2}{\mathrm{n}_2}ight)}$
$$\Rightarrow $$ $\frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{Z}_1}{\mathrm{n}_1} 	imes \frac{\mathrm{n}_2}{\mathrm{Z}_2}$ .....(vi)
$$\Rightarrow $$ $$ \frac{\mathrm{v}_1}{\mathrm{v}_2}=\left[\left(\frac{\mathrm{n}_1}{\mathrm{n}_2}ight)^2ight] \cdot\left(\frac{\mathrm{n}_2}{\mathrm{n}_1}ight)=\frac{\mathrm{n}_1}{\mathrm{n}_2} $$
It is given that the speed of first electron is $\mathrm{v}_1=3 	imes 10^5 \mathrm{~m} / \mathrm{s}$ and that of electron in second hydrogen like atom is $\mathrm{v}_2=2.5 	imes 10^5 \mathrm{~m} / \mathrm{s}$
$$ \frac{n_1}{n_2}=\frac{3 	imes 10^5}{2.5 	imes 10^5} $$
$$\Rightarrow $$ $\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{2.5}$
$$\Rightarrow $$ $\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{6}{5}$
So, the smallest possible energy states in two atoms are 6 and 5.
Hence, the correct option is (D).

Answer

8 and 10

Answer

10 and 12

Answer

9 and 8

Answer

6 and 5

Atoms & Nuclei

Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 \times 10^5 \mathrm{~m} / \mathrm{s} and 2.5 \times 10^5 \mathrm{~m} / \mathrm{s} respectively. If the radii of these orbits are nearly same then the possible order of energy states are \_\_\_\_ respectively.

Choose the correct answer: