The spectral lines of the Lyman series are emitted when electrons in a hydrogen atom transition to the n=1 energy level. Which of the following transitions would produce the photon with the SHORTEST wavelength in the Lyman series?

The energy of an electron in $n\,\,th$ orbit of hydrogen atom is $- 13.6/{n^2}\,\,{\rm{eV}}$. Energy required to excite the electron from the first orbit to the third orbit is

Energy required for the electron excitation in $L{i^{ + + }}$ from the first to the third Bohr orbit is

An electron in an atom has a total energy of $-5.4$ eV. What are its kinetic and potential energies, respectively?

Which of the following correctly relates the kinetic energy (KE), potential energy (PE), and total energy (E) of an electron in a Bohr orbit of hydrogen?

If the shortest wavelength of the Balmer series of hydrogen is $\lambda$, then the longest wavelength in the Lyman series is?

In a beryllium atom, if ${a_0}$ be the radius of the first orbit, then the radius of the second orbit will be will be in general

The radius of electron’s second stationary orbit in Bohr’s atom is $R$. The radius of the third orbit will be

According to Bohr’s theory the radius of electron in an orbit described by principle quantum number $n$ and atomic number $Z$ is proportional to

When hydrogen atom is in its first excited level, its radius is its ground state radius

Which of the following transitions in the hydrogen atom corresponds to the shortest wavelength in the Lyman series?