(B) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by $r_n = a_0 \times \frac{n^2}{Z}$, where $a_0$ is the Bohr radius $(52.9 \t

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

(B) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by $r_n = a_0 \times \frac{n^2}{Z}$, where $a_0$ is the Bohr radius $(52.9 \text{ pm})$.For $H$ atom $(Z=1)$: $x = r_3 - r_2 = a_0 \times (3^2 - 2^2) / 1 = a_0 \times (9 - 4) = 5a_0$.For $Li^{2+}$ ion $(Z=3)$: $y = r_4 - r_3 = a_0 \times (4^2 - 3^2) / 3 = a_0 \times (16 - 9) / 3 = \frac{7}{3}a_0$.The ratio $y:x = (\frac{7}{3}a_0) : (5a_0) = \frac{7}{3} : 5 = 7 : 15$.

Class 11 Chemistry Structure of Atom Atomic models and Planck's quantum theory

Medium

The difference between the radii of $3^{rd}$ and $2^{nd}$ orbit of $H$ atom is $x \text{ pm}$. The difference between the radii of $4^{th}$ and $3^{rd}$ orbit of $Li^{2+}$ ion is $y \text{ pm}$. The ratio $y:x$ is equal to:

AP EAMCET 2025 All AP EAMCET

A
$15:7$
B
$7:15$
C
$3:1$
D
$1:3$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

Structure of Atom

Subtopic

Atomic models and Planck's quantum theory

Previous Year

AP EAMCET 2025 Paper All AP EAMCET years →

Calculate the wavelength for the emission transition if it starts from the orbit having radius $1.3225 \, nm$ and ends at $211.6 \, pm$. Name the series to which this transition belongs and the region of the spectrum.

Difficult

View Solution

The value of Planck's constant is $6.63 \times 10^{-34} \ J \ s$. The velocity of light is $3.0 \times 10^8 \ m \ s^{-1}$. Which value is closest to the wavelength in nanometres of a quantum of light with frequency of $8 \times 10^{15} \ s^{-1}$?

MediumAIPMT 2003

View Solution

Assertion $:-$ Emitted radiation will fall in the visible region when an electron jumps from $n=4$ to $n=2$ in a hydrogen atom.
Reason $:-$ The frequency of radiation of the Lyman series belongs to the visible region for a hydrogen atom.

Easy

View Solution

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $n=2 \rightarrow n=3$ and $n=4 \rightarrow n=6$ transitions,respectively,is

MediumNEET 2025

View Solution

Assertion : For Balmer series of hydrogen spectrum,the value $n_1 = 2$ and $n_2 = 3, 4, 5...$
Reason : The value of $n$ for a line in Balmer series of hydrogen spectrum having the highest wavelength is $4$ and $6$.

MediumAIIMS 2002

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The difference between the radii of $3^{rd}$ and $2^{nd}$ orbit of $H$ atom is $x \text{ pm}$. The difference between the radii of $4^{th}$ and $3^{rd}$ orbit of $Li^{2+}$ ion is $y \text{ pm}$. The ratio $y:x$ is equal to:

Similar Questions

Calculate the wavelength for the emission transition if it starts from the orbit having radius $1.3225 \, nm$ and ends at $211.6 \, pm$. Name the series to which this transition belongs and the region of the spectrum.

The value of Planck's constant is $6.63 \times 10^{-34} \ J \ s$. The velocity of light is $3.0 \times 10^8 \ m \ s^{-1}$. Which value is closest to the wavelength in nanometres of a quantum of light with frequency of $8 \times 10^{15} \ s^{-1}$?

Assertion $:-$ Emitted radiation will fall in the visible region when an electron jumps from $n=4$ to $n=2$ in a hydrogen atom.Reason $:-$ The frequency of radiation of the Lyman series belongs to the visible region for a hydrogen atom.

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $n=2 \rightarrow n=3$ and $n=4 \rightarrow n=6$ transitions,respectively,is

Assertion : For Balmer series of hydrogen spectrum,the value $n_1 = 2$ and $n_2 = 3, 4, 5...$ Reason : The value of $n$ for a line in Balmer series of hydrogen spectrum having the highest wavelength is $4$ and $6$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $:-$ Emitted radiation will fall in the visible region when an electron jumps from $n=4$ to $n=2$ in a hydrogen atom.
Reason $:-$ The frequency of radiation of the Lyman series belongs to the visible region for a hydrogen atom.

Assertion : For Balmer series of hydrogen spectrum,the value $n_1 = 2$ and $n_2 = 3, 4, 5...$
Reason : The value of $n$ for a line in Balmer series of hydrogen spectrum having the highest wavelength is $4$ and $6$.