Bohr model and hydrogen spectrum

$\dfrac x{x+4}$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its

(i) third permitted energy level to the second level and

(ii) the highest permitted energy level to the second permitted level.

The value of $x$ will be ______.

In the hydrogen spectrum, $\lambda$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "$a\lambda$" between the wavelength of 3$^{\text{rd}}$ transition line of Paschen series and that of 2$^{\text{nd}}$ transition line of Balmer series where $a=$ ______.

Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.

Using Bohr model of quantization of angular momentum the relation between the radius 'r' of the n$^{th}$ allowed orbit of quantum number 'n' for an electron in hydrogen atom is:

Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom?

In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E.

The transitions A, B and C respectively represent :

A particle of mass $m$ moves in circular orbits with potential energy $V(r) = Fr$, where $F$ is a positive constant and $r$ is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle's orbit is denoted by $R$ and its speed and energy are denoted by $v$ and $E$, respectively, then for the $n^{\text{th}}$ orbit (here $h$ is the Planck's constant)

A free hydrogen atom after absorbing a photon of wavelength $\lambda_a$ gets excited from the state $n = 1$ to the state $n = 4$. Immediately after that the electron jumps to $n = m$ state by emitting a photon of wavelength $\lambda_e$. Let the change in momentum of atom due to the absorption and the emission are $\Delta p_a$ and $\Delta p_e$, respectively. If $\lambda_a/\lambda_e = \frac15$, which of the option(s) is/are correct?

[Use $hc = 1242$ eV nm; 1 nm $= 10^{-9}$ m, $h$ and $c$ are Planck's constant and speed of light, respectively]

Consider a hydrogen-like ionized atom with atomic number $Z$ with a single electron. In the emission spectrum of this atom, the photon emitted in the $n = 2$ to $n = 1$ transition has energy 74.8 eV higher than the photon emitted in the $n = 3$ to $n = 2$ transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of $Z$ is __________.

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $n_i$ to another with quantum number $n_f$. $V_i$ and $V_f$ are respectively the initial and final potential energies of the electron. If $\frac{V_i}{V_f} = 6.25$, then the smallest possible $n_f$ is

Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths $r=\lambda_1/\lambda_2$, is given by :

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$^{\text{th}}$ orbit is proportional to : (n = principal quantum number)

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

Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge $Ze$ are defined by their principal quantum number $n$, where $n\gg1$. Which of the following statement(s) is(are) true?

A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å. Taking $hc/e=1.237\times10^{-6}$ eV m and the ground state energy of hydrogen atom as $-13.6$ eV, the number of lines present in the emission spectrum is

A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :

A neutron moving with a speed '$v$' makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place is :

Consider a hydrogen atom with its electron in the $n^{th}$ orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of $n$ is ($hc=1242$ eV nm)

An electron in an excited state of $\ce{Li^{2+}}$ ion has angular momentum $3h/2\pi$. The de Broglie wavelength of the electron in this state is $p\pi a_0$ (where $a_0$ is the Bohr radius). The value of $p$ is

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