S. L. Arora Solutions for Chapter: Atoms, Exercise 1: Problems For Practice

The second member of Lyman's series in hydrogen spectrum has a wavelength of $5400 {\mathrm{A}}^0$. Calculate the wavelength of the first member.

A photon of energy $12.09 \mathrm{eV}$ is absorbed by an electron in ground state of a hydrogen atoms. What will be the energy level of electron? The energy of electron in the ground state of hydrogen atom is $-13.6 \mathrm{eV}.$

The period of revolution of the electron in the third orbit in a hydrogen atom is $4.132\times {10}^{-15} \mathrm{s}$. Hence find the period in the $5$ th Bohr orbit.

The electron, in a hydrogen atom, is in its second excited state. Calculate the wavelength of the lines in the Lyman series, that can be emitted through the permissible transitions of this electron.(Given the value of Rydberg constant, $\left.R=1.1\times {10}^7 {\mathrm{m}}^{-1}\right).$

Show that the shortest wavelength lines in Lyman, Balmer and Paschen series have their wavelengths in the ratio $1:4:9$.

The short wavelength limit for the Lyman series of the hydrogen spectrum is $913.4 {\mathrm{A}}^0$. Calculate the short wavelength limit for Balmer series of hydrogen spectrum.

Find the ratio between the wavelengths of the 'most energetic' spectral lines in the Balmer and Paschen series of the hydrogen spectrum.

Find out the wavelength of the electron orbiting in the ground state of hydrogen atom.