Atoms Questions

The ground state energy of hydrogen atom is $-13.6\text{eV}$. What is the potential energy of the electron in this state?

The ionization potential of the hydrogen atom is $13.6\text{V}$. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy $12.1\text{eV}$. According to Bohr’s theory, the spectral lines emitted by hydrogen will be:

In a discharge tube ionization of enclosed gas is produced due to collisions between:

In a Rutherford scattering experiment when a projectile of charge $z_1$ and mass $M_1$ approaches a target nucleus of charge $z_2$ and mass $M_2$, the distance of closest approach is $r_0$. The energy of the projectile is:

The ionization energy of the electron in the hydrogen atom in its ground state is $13.6\text{eV}$. The atoms are excited to higher energy levels to emit radiations of $6$ wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between:

Transition $n=3$ to $n=1$ results in ultraviolet radiation. Infrared radiation will be obtained in:

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be:

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths ${\lambda }_1:{\lambda }_2$ emitted in the two cases is:

Given the value of Rydberg constant is $10^7{\text{m}}^{-1}$, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

When an $\alpha$ -particle of mass $m$ moving with velocity $v$ bombards on a heavy nucleus of charge $Ze$, its distance of closest approach from the nucleus depends on $m$ as:

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength $\lambda$. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be: