The ratio of the radius of the second orbit o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The radius of an electron $(r_n)$ in any orbit is given by the formula: $r_n = 0.529 	imes \frac{n^2}{Z} \ \mathring{A}$.
For $Li^{2+}$: $n = 2$

Vedclass · Accepted Answer

$16/27$

Vedclass · Answer

(D) The radius of an electron $(r_n)$ in any orbit is given by the formula: $r_n = 0.529 	imes \frac{n^2}{Z} \ \mathring{A}$.
For $Li^{2+}$: $n = 2$, $Z = 3$.
For $Be^{3+}$: $n = 3$, $Z = 4$.
The ratio is given by: $\frac{r_{Li^{2+}}}{r_{Be^{3+}}} = \frac{n_1^2 / Z_1}{n_2^2 / Z_2} = \frac{n_1^2}{Z_1} 	imes \frac{Z_2}{n_2^2}$.
Substituting the values: $\frac{2^2}{3} 	imes \frac{4}{3^2} = \frac{4}{3} 	imes \frac{4}{9} = \frac{16}{27}$.
Thus, the correct option is $(D)$.

The ratio of the radius of the second orbit of $Li^{2+}$ to that of the third orbit of $Be^{3+}$ is:

Similar Questions

The ground state energy of $H$ is equal to

When a hydrogen atom in an excited state undergoes electronic transitions to the ground state,$5$ spectral lines are observed in the ultraviolet region. How many spectral lines will be observed in the infrared region due to transitions from the same excited state?

The emission spectrum of a gaseous atom is line-based and not continuous. Why?

The energy of the electron at infinite distance from the nucleus in Bohr’s model is taken as:

The four lowest energy levels of the $H$ atom are shown in the figure. The number of emission lines will be ...... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module