How radius depends upon n in the Bohrs hydrogen atom model nprincipalquantumnumber (2024)

To calculate the size of a hydrogen anion using the Bohr model, we assume that its two electrons move in an orbit such that they are always on diametrically opposite sides of the nucleus. With each electron having the angular momentum,$♄=\frac{h}{2\pi },$and taking electron interaction into account the radius of the orbit in terms of the Bohr radius of a hydrogen atom $a_B=\frac{4\pi {\epsilon }_0{h}^2}{m{e}^2}$ is

In a hydrogen like atom electron makes transition from an energy level with quantum number $\text{n}$ to another with quantum number $\left(\text{n}-1\right)$. If $\text{n >> 1}$, the frequency of radiation emitted is proportional to :

The magnitude of acceleration of the electron in the $n^{\text{th}}$orbit of hydrogen atom is $a_{\mathrm{H}}$ and that of singly ionised helium atom is $a_{\mathrm{He}}$. The ratio $a_{\mathrm{H}}:a_{\mathrm{He}}$ is,

If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength$\lambda$. When it jumps from the fourth orbit to the third orbit, the corresponding wavelength of the photon will be

A hydrogen atom is in the ground state. Then to get six lines in an emission spectrum, wavelength of the incident radiation should be

Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths ${\lambda }_N,{\lambda }_A$ respectively. The ratio $\frac{{\lambda }_N}{{\lambda }_A}$ is closest to:

A particle of mass,$m$ moves around the origin in a potential,$\frac{1}{2}m{\omega }^2{r}^2$,where $r$ is the distance from the origin.Applying the Bohr model in this case, the radius of the particle in its $n^{\text{th}}$ orbit in terms of $a=\sqrt{\frac{h}{\left(2\pi m\omega \right)}}$ is,

According to Bohr's theory, the time averaged magnetic field at the centre (i.e., nucleus) of a hydrogen atom due to the motion of electrons in the $n^{th}$ orbit is proportional to: ($n=$ principal quantum number)

As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion

A particle of mass $m$ moves in a circular orbit in a central potential field $U\left(r\right)=\frac{1}{2}k{r}^2.$ If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number $n$ as:

The energy required to remove the electron from a singly ionized Helium atom is $2.2$ times the energy required to remove an electron from helium atom. The total energy required to ionize the Helium atom completely is close to

An electron makes a transition from an excited state to the ground state of a hydrogen-like atom. Then

An electron with kinetic energy $E$ collides with a hydrogen atom in the ground state. The collision will be elastic

Consider thirdorbit of ${\mathrm{He}}^{+}$ (helium), using non-relativistic approach, the speed of electron in this orbit will be given constant $K{\text{=9×10}}^{\text{9}}$,$Z=2$ and $h$ (Planck's Constant)$\text{=6}\text{.6}\times {\text{10}}^{-34}\text{Js}$

An electron in a hydrogen atom jumps from second Bohr orbit to ground state and the energy difference of the two states is radiated in the form of photons. These are then allowed to fall on a metal surface having a work-function equal to$\text{4.2}\mathrm{eV}$, then the stopping potential is [Energy of electron in ${\mathrm{n}}^{\mathrm{th}}$ orbit $=-\frac{13.6}{{\mathrm{n}}^2}\mathrm{eV}$]

The wavelength of the first Balmer line caused by a transition from the $n=3$ level to the $n=2$ level in hydrogen is ${\lambda }_1.$ The wavelength of the line caused by an electronic transition from $n=5$ to $n=3$ is

The acceleration of an electron in the first orbit of the hydrogen atom ($n=1$) is :

In a Frank - Hertz experiment, an electron of energy $5.6\mathrm{eV}$passes through mercury vapour and emerges with an energy $0.7\mathrm{eV}.$ The minimum wavelength of photons emitted by mercury atoms is close to:

The ratio of radii of the first three Bohr's orbits is

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is