Physics M.C.Q (1 Marks)

Energy of photon $=\frac{\text{hc}}{\gamma}$
Hence, a photon with longer wavelenght will carry less energy.

Atomic No $=$ no of proton $= 8$
Oxygen has atomic number $8$. So, element $X$

Niels Bohr introduced the atomic Hydrogen model in $1913.$
Neils Bohr developed the Bohr model of the atom, in which he proposed that energy levels of electrons are discrete and that the electrons revolve in stable orbits around the atomic nucleus but can jump from one energy level $($or orbit$)$ to another.

Radius is directly proportional to square of $n.$
Therefore, ratio of principal quantum number is $1 : 3.$
Hence, ratio of frequencies is $27 : 1.$

The energies of the photons emitted can be expressed as follows,

de $–$ Broglie hypothesis did modify Bohr’s second postulate.
This postulate of Bohr regarding the quantization of the angular momentum of an electron was further explained by Louis de Broglie.
According to de $–$ Broglie, a moving electron in its circular orbit behaves like a particle$-$wave.

The word atom is derived from Greek word, 'atomos'. In Greek, the prefix $"a"$ means "not" and the word "tomos" means cut. Our word atom therefore comes from atomos, a Greek word meaning uncuttable.

For a hydrogen-like ion with $Z$ protons in the nucleus, the radius of the nth state is given by,
$\text{r}_\text{n}=\frac{\text{n}^2\text{a}_0}{\text{Z}}$
Here,
$a_0 = 0.53\ pm$
For lithium, $Z = 3$ Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.

The energy of characteristic $X-$ray is a consequence of transition in target atoms.
They cause emission or absorption of electromagnetic radiation.
The other options are not responsible for the energy of characteristic $X-$rays.

The valence electron in an alkali metal is an s-electron.
Generally, they make up Group $1$ of the periodic table.
The different examples that come under this category are lithium, potassium, and francium.

The radius of an electron orbit in a hydrogen atom is of the order of $ 10^{-11} \mathrm{~m} $.
It is equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state.

The ionisation energy of a hydrogen like ion of atomic number $Z$ is given by,
$\text{V}=(13.6\text{eV})\times\text{Z}^2$
Thus, the atomic number of ion $A$ is greater than that of B$(Z_A > Z_B)$.
The radius of the orbit is inversely proportional to the atomic number of the ion.
$\therefore\ \text{r}_\text{A}>\text{r}_\text{B}$
The speed of electron is directly proportional to the atomic number.
Therefore, the speed of the electron in the orbit of $A$ will be more than that in $B.$
Thus, $u_A > u_B$ is correct.
The total energy of the atom is given by,
$\text{E}=-\frac{\text{mZ}^2\text{e}^2}{8\in_0\text{h}^2\text{n}^2}$
As the energy is directly proportional to $Z^2$, the energy of $A$ will be less than that of $B,$ i.e. $E_A < E_B$.
The orbital angular momentum of the electron is independent of the atomic number. Therefore, the relation $L_A > L_B$ is invalid.

Molybdenum is used as a target in Coolidge tube for product.
$X-$ ray tube is an energy converter and it’s a device made of cathode and anode.
An electrical current flows through the tube from cathode to anode, when electron undergoes a energy loss, which results in generation of $x -$rays.
Anode is a component in which $x-$rays are produced. It has $2$ primary functions. To convert electricity to $x-$ray
To dissipate heat in the process
For this, material such as molybdenum and tungsten are used with high atomic number which has good heat storage capacity and low rate of evaporation.

For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}_1}-\frac{1}{\text{n}_2}\Big)$
Here, $R$ is the Rydberg constant.
For the transition from $n = 5$ to $n = 4$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{4^2}-\frac{1}{5^2}\Big)$
$\lambda=\frac{400}{9\text{RZ}^2}$
For the transition from $n = 4$ to $n = 3$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{3^2}-\frac{1}{4^2}\Big)$
$\lambda=\frac{144}{7\text{RZ}^2}$
For the transition from $n = 3$ to $n = 2$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)$
$\lambda=\frac{36}{5\text{RZ}^2}$
For the transition form $n = 2$ to $n = 1$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)$
$\lambda=\frac{2}{\text{RZ}^2}$
From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from $n = 2$ to $n = 1$ will be minimum.