l represents the subshell and L represents shell.

s-orbital.

'1' signifies the secondary quantum number.
'L' signifies second energy level (n = 2).

The metals with low ionisation enthalpies are used in photoelectric cells. Caesium (Cs), an alkali metal belonging to group 1 is the most commonly used metal.

A quantum is a bundle of energy of a definite magnitude (E = hv) and it may be from any source. However, a photon is quantum of energy associated with light only.

Both have identical shape consisting of four lobes.

Cosmic rays> X-rays> radio waves.

Relation:\(\lambda =\frac { h }{ mV } \)
The relation is known as de Broglie's relationship.

Let the number of electrons in an ion = x
Number of neutrons, n \(=x+\frac { 11.1 }{ 100 } x=1.111x\)
(As the number of neutrons are 11.1% more than the number of electrons.)
Since, the ion carries -1 charge, the number of protons, p = x - 1
we know that, mass number = n + p = 37
or  1.111 x + x - 1 = 37 or 2.111 x = 37 + 1 = 38
\(x=\frac { 38 }{ 2.111 } =18.0009 \ \approx 18\)
Number of protons = atomic number = 18 - 1 = 17
Therefore, the symbol of the ion is \(_{ 17 }^{ 37 }{ CI^{ - } }\)

\({ H }_{ 2 }=_{ 1 }H+_{ 1 }H=1+1=2{ e }^{ - }\)
\({ H }_{ 2 }^{ + }=2-1=1e^{ - }\)
\( { O }_{ 2 }=_{ 8 }O+_{ 8 }O=8+8=16{ e }^{ - }\)
\( { O }_{ 2 }^{ + }=16-1=15e^{ - }\)

\( \lambda=25.0 \mu m=25.0 \times 10^{-6} \mathrm{~m} \)
\( v=\frac{c}{\lambda}=\frac{3.0 \times 10^8}{25.0 \times 10^{-6}}=1.2 \times 10^{13}\)
Energy of photon,
\( E=h v=6.626 \times 10^{-34} \times 1.2 \times 10^{13} \)
\( =7.96 \times 10^{-21} \mathrm{~J}\)

1.67 x 10⁸ m^-1

From the expression of energy of one photon (E),
E=hc/λ
Where, λ = wavelength of radiation
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of E:
E\(=\frac{(6.626×10^{−34}Js)(3×108ms^{−1})}{(600×10^{−9}m)}\)=3.313×10⁻¹⁹J
Energy of one photon = 3.313 × 10^–19 J
Number of photons received with 3.15 × 10^–18 J energy
\(= \frac{3.15×10^{−18}J}{3.313×10^{−19}J}\)=9.5≈10

Energy (E) of one photon is given by E = hv
h = 6.626 x 10^-34 Js.
v = 5 x 10¹⁴ s^-1 (given)
E = (6.626 x 10^-34 Js) x (5 x 10¹⁴s^-1) = 3.313 x 10^-19 J
Energy of one mole of photons
= (3.313 x 10^-19 J) x (6.022 x 10²³mol^-1) = 199.51 KJ mol^-1

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.