Hydrogen.

In the construction of electron microscope used for the measurement of objects of very small size.

Ground state means the lowest energy state. When the electrons absorb energy and jump to outer orbits, this state is called excited state.

As Fe³⁺ contains 5 unpaired electrons while Fe²⁺ contains only 4 unpaired electrons. Fe³⁺ is more paramagnetic.

The Balmer series lies in the visible region.

Energy, \(E=\frac { hc }{ \lambda } \)
\(\lambda =0.50\overset { \circ }{ A } \ =0.50\times { 10 }^{ -10 }m\)
\(E=\frac { 6.626\times { 10 }^{ -34 } \ Js\times 3\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 0.50\times { 10 }^{ -10 }m } \)
= 39.756 x 10^-16 J
= 3.975 x 10^-15 J

(i) Energy, E = hv
[h = Planck's constant = 6.626 x 10^-34Js v = 3 x 10¹⁵ Hz = 3 x 10¹⁵Cps]
Energy, E = 6.626 x 10^-34 Js x 3 x 10¹⁵ s^-1 
= 19.878 x 10^-19 J = 1.9878 x 10^-18 J
(ii) Energy, \(E=\frac { hc }{ \lambda } \)
\(\lambda =0.50\overset { \circ }{ A } \ =0.50\times { 10 }^{ -10 }m\)
\(E=\frac { 6.626\times { 10 }^{ -34 } \ Js\times 3\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 0.50\times { 10 }^{ -10 }m } \)
= 39.756 x 10^-16 J
= 3.975 x 10^-15 J

In case of positive rays, the ions remaining after the loss of electrons might have the same magnitude of charge but different masses. Hence they will have different charge to mass ratio. Cathode rays are made up of electrons and all electrons have same charge to mass ratio. That's why charge to mass ratio of all cathode rays is same.

As A = n + p
p = A - n = 13 - 7 = 6
Hence, atomic number, z = p = 6

Mass number is a whole number because it is the sum of the number of protons and number of neutrons whereas atomic mass is fractional because it is the average relative mass.

Neutron is a neutral practice. Hence it will not be deflected on passing through an electric field.

Charge on the oil drop = 1.282 ×10^–18C
Charge on one electron = 1.6022 × 10^–19C
∴ Number of electrons present on the oil drop
\(=\frac{−1.282 × 10^{−18}C}{−1.6022×10^{−19}C}\)=0.800×10=8.0 electrons

Cathode rays originate from the cathode whereas anode rays are not obtained from the anode. They are produced from the gaseous atoms by knock out of the electrone with high speed cathode rays.

Step I. Calculation of the velocity of electron
Kinetic energy = 1/2 mv² = 4.55 x 10^-25J = 4.55 x 10^-25 kg m² S^-2
or v² = \(\frac { 2\times KE }{ m } =\frac { 2\times (4.55\times10^{ -25 }kgm^{ 2 }s^{ -2 }) }{ (3x10^{ 6 }ms^{ -1 }) } \)=10⁶ m² S^-2
or Velocity (v) = (10⁶ m² S^-2 )^1/2 = 10³ ms^-1
Step II. Calculation of the momentum of the electron Momentum of electron = mv = (9.1 x 10^-31 kg) x (10³m s^-l) = 9.1 x 10^-28 kg m^-1
Step III. Calculation of the wavelength of the electron According to de Broglie equation:
\(\lambda \frac { h }{ mv } =\frac { (6.626\times10^{ -34 }kgm^{ 2 }s^{ -1 }) }{ (9.1x10^{ -31 }kg)\times (10^{ 3 }ms^{ -l }) } \)
= 0.728 x 10-6m = 7.28 x 10-7 m

According to de Broglie equation,\(\lambda =\frac { h }{ mv } \)
Mass of electron = 9.1 x 10^-31 kg; Planck's constant = 6.626 x 10^-34 kgm²s^-1
Velocity of electron = 1% of speed of light = 3.0 x 10⁸ x 0.01 = 3 x 10⁶ me-1
Wavelength of electron (\(\lambda \)) =\(\frac { h }{ mv } =\frac { (6.626x10^{ -34 }kgm^{ 2 }s^{ -1 }) }{ (9.1x10^{ -31 }kg)x(3x10^{ 6 }ms^{ -1 }) } \)
= 2.43 x 10^-10 m.

\(\lambda _{ O_{ 2 } }/\lambda _{ SO_{ 2 } }=2 \ (because \ \lambda =\frac { h }{ mv } ,i.e.\lambda \propto \frac { 1 }{ m } \) and mass of SO₂ molecule viz.64 u is double than that of O₂ molecule viz.32 u).

(a) \(_{ 92 }^{ 233 }{ U }\)
(b) \(^{ 9 }{ Be }\)

Atomic Number (Z): The atomic number of an element is equal to the number of protons present inside the nucleus of its atoms. Since, an isolated atom has no net charge on it, in neutral atoms, the total number of electrons is equal to its atomic number. 
Atomic number (Z) = Number of protons in the nucleus of an atom
= Number of electrons in the neutral atoms
Mass Number (A): The sum of the number of neutrons and protons in the nucleus of an atom is called its mass number. Mass number is denoted by A. Thus, for an atom, Mass number (A) = Number of protons (P) + Number of neutrons (n) A = P + n
Neutron: It is neutral particle. It is present in the nucleus of an atom. Expect hydrogen (which contains only one electron and one proton but no neutron), the atoms of all other elements including isotopes of hydrogen contain all the three fundamental particles called neutron, proton and electron.
The relation between mass number, Atomic no. and no. of neutrons is given by the equation:
A = Z+n I
Where A = Mass number
Z = Atomic number
n = Number of neutrons in the nucleus.

Given, velocity of beam of helium atoms = 2.0\(\times \)10³m sec^-1
Mass of helium atom = \(\frac { 4 }{ 6.022\times { 10 }^{ 23 } } \)
= \(6.64\times { 10 }^{ -24 }g=6.64\times { 10 }^{ -27 }kg\)
According to de-Broglie equation, \(\lambda =\frac { h }{ mv } \)
\(=\frac { 6.626\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 } }{ (6.64\times { 10 }^{ -27 }kg)\times (2.0\times { 10 }^{ 3 }{ ms }^{ -1 }) } \)
\(= 4.99\times { 10 }^{ -11 }m=49.9pm \ [2]\)

Since \({ n }_{ i }=5\) and \({ n }_{ f }=2\). this transition gives rise to a spectral line in the visible region of the Balmer series. From equation
\(\triangle E=2.18\times 10^{ -18 }J\left[ \frac { 1 }{ { n }^{ 2 }_{ i } } -\frac { 1 }{ { n }_{ f }^{ 2 } } \right] \)
\(=2.18\times 10^{ -18 }J\left[ \frac { 1 }{ { { 5 }^{ 2 } } } -\frac { 1 }{ { 2 }^{ 2 } } \right] =-4.58\times 10^{ -19 }J\)
It is an emission energy.
The frequency of the photon (taking energy in terms of magnitude) is given by
\(v=\frac { \triangle E }{ h } =\frac { 4.58\times 10^{ -19 }J }{ 6.626\times 10^{ -34 }Js }\)
\(=6.91\times 10^{ 14 }Hz\)
\(\\ \lambda =\frac { c }{ v } =\frac { 3.0\times 10^{ 8 }ms^{ -1 } }{ 6.91\times 10^{ 14 }Hz } =434nm\)

(a) Calculation of wavenumber \((\bar{V})\)
 λ=5800Å = 5800 × 10^–8 cm = 5800 × 10^–10 m
\( \overline { V } =\frac { 1 }{ \lambda } =\frac { 1 }{ 5800\times { 10 }^{ -10 }m }\)
\(=1.741\times { 10 }^{ 6 }m=1.724\times { 10 }^{ 4 }{ cm }^{ -1 }\)
(b) Calculation of the frequency (v)
\(v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 5800\times { 10 }^{ -10 }m }\)
\( =5.172\times { 10 }^{ 14 }{ s }^{ -1 }\)