Physics M.C.Q (1 Marks)

Atomic number of any element is equal to number of proton or number of electron
Atomic Number $=$ no. of proton $=$ no of electron.

The radius of the $\mathrm{n}^{\text {th }}$ orbit in one electron system is given by,
$\mathrm{r}_{\mathrm{n}}=\frac{\mathrm{n}^2 \mathrm{a}_0}{\mathrm{z}}$
Here, $a_0=53 \mathrm{pm}$
For the shortest orbit, $\mathrm{n}=1$
For hydrogen, $Z=1$
$\therefore$ Radius of the first state of hydrogen atom $=53 \mathrm{pm}$
For deuterium, $Z=1$
$\therefore$ Radius of the first state of deuterium atom $=53 \mathrm{pm}$
For $\mathrm{He}^{+}, \mathrm{Z}=2$
$\therefore$ Radius of $\mathrm{He}^{+}$atom $=\frac{53}{2} \mathrm{pm}=26.5 \mathrm{pm}$
For $\mathrm{Li}^{++}, \mathrm{Z}=3$
$\therefore$ Radius of $\mathrm{Li}^{++}$atom $=\frac{53}{3} \mathrm{pm}=17.66 \approx 18 \mathrm{pm}$
The given one$-$electron system having radius of the shortest orbit to be $18\ pm$ may be $\mathrm{Li}^{++}$

$J.J$. Thomson proposed plum pudding model in $1904$. Ernest Rutherford introduced the planetary model of an atom in $1911$ which was modified by Neils Bohr in $1913$ as Bohr model of an atom.
Order in which the atomic models were advanced: Plum pudding, planetary, Bohr.

In a Cathode ray tube electrons comes out from cathode. After cathode, anode is placed, which as being $+$ ively charged, accelerates electrons and provide them kinetic energy.

When an alpha particle is emitted from a radioactive source or substance, its atomic number decreases by $2$ and its atomic mass decreases by $4$, which is same as that of helium ion.

de $–$ Broglie concluded his modification of Bohr’s second postulate by stating that the wavelengths of matter waves can be quantized.
This implies that the electrons can exist in those orbits which had a complete set of several wavelengths.

The emission spectrum of $\mathrm{CO}_2$​ gas that has been studied by the electron beam excitation method is a band spectrum.

The first postulate of Bhor theory is that the orbital momentum of the electron is quantized ie, $L = mvr = nh$ where $h$ is Drac constant.

The minimum energy required to excite a hydrogen atom from its ground state to $1^{\text {st }}$ excited state is approximately $10\ eV.$ As the incident electron energy is not sufficient for excitation of the hydrogen atom so electron will not get absorbed in the hydrogen atom so it can not be an inelastic collision. Also this collision can not be partially elastic because in an partially elastic collision, there is a net loss on kinetic energy. If the energy is lost then corresponding amount of heat should have been produced but it is not so which implies that the collision is completely elastic.

Since the brightness or intensity of the display depends on the number of electrons that strike the screen, the control grid is used to control the brightness of the $\text{CRT.}$

The total energy of a hydrogen$-$like ion, having $Z$ protons in its nucleus, is given by,
$\text{E}=-\frac{13.6\text{Z}^2}{\text{n}^2}\text{eV}$
Here, $n =$ Principal quantum number.
For ground state, $n = 1$
$\therefore$ Total energy, $E = -13.6Z^2\ eV$
For hydrogen, $Z = 1$
$\therefore$ Total energy, $E = -13.6\ eV$
For deuterium, $Z = 1$
$\therefore$ Total energy, $e = -13.6\ eV$
For $He^+, Z = 2$
$\therefore$ Total energy $E = -13.6 \times 2^2 = -54.4eV$
For $Li^{++}$,
$Z = 3$
$\therefore$ Total energy, $E = -13.6 \times 3^2 = -122.4\ eV$
Hence, the ion having an energy of $-54.4\ eV$ in its ground state may be $He^+$

The electrons are accelerated by a second anode at high potential, more than $500V$

Niels Bohr calculated the energies that a hydrogen atom would have in each of its energy levels, based on the wavelength of the spectral lines.
Then he found out that there are four spectral lines for hydrogen, namely, Lyman, Balmer, Paschen, and Brackett series.
The Lyman series lies in the $UV$ region, whereas the Balmer series lies in the visible region, and the last two lie in the infrared region.

Values of Principle quantum number are $1, 2, 3, 4.....8. 0$ is not a Principle quantum number.
Only odd numbers are not Principle quantum numbers either.
Odd numbers, as well as even numbers, are Principle quantum number except $0$ and negative integers.