What is the radius ratio for the 2^{nd} orbit | vedclass

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

Question

(B) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = a_0 	imes \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is

Vedclass · Accepted Answer

$16 : 27$

Vedclass · Answer

(B) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = a_0 	imes \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is the orbit number,and $Z$ is the atomic number.For $Li^{+2}$ ion $(Z=3)$: $n_1 = 2$,$Z_1 = 3$. So,$r_1 = a_0 	imes \frac{2^2}{3} = a_0 	imes \frac{4}{3}$.For $Be^{+3}$ ion $(Z=4)$: $n_2 = 3$,$Z_2 = 4$. So,$r_2 = a_0 	imes \frac{3^2}{4} = a_0 	imes \frac{9}{4}$.The ratio is $\frac{r_1}{r_2} = \frac{4/3}{9/4} = \frac{4}{3} 	imes \frac{4}{9} = \frac{16}{27}$.

What is the radius ratio for the $2^{nd}$ orbit of $Li^{+2}$ ion to the $3^{rd}$ orbit of $Be^{+3}$ ion?

Similar Questions

Which of the following relations is correct for the energy of the ground state of hydrogen-like species?

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{h^{2}}{xma_{0}^{2}}$. The value of $10x$ is ........ . ($a_{0}$ is radius of Bohr's orbit) (Nearest integer) [Given : $\pi=3.14$]

The surface of sodium metal is irradiated with radiation of wavelength $x$ nm. The kinetic energy of ejected electrons is $2.8 \times 10^{-20}$ $J$. The work function of sodium is $2.3$ eV. The value of $x$ is . . . . . . $\times 10^2$ nm.

The ratio of the highest to the lowest wavelength of Lyman series is

Calculate the energy of an electron in the first four orbits of a hydrogen atom in $kJ \, mol^{-1}$ (Given: $N_A = 6.02 \times 10^{23} \, mol^{-1}$,Rydberg constant $R_H = 2.18 \times 10^{-18} \, J$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module