K_max = eV_s = e(1.5V) = 1.5eV
= 1.5 x 1.6 x 10^-19J
= 2.4 x 10^-19J

Principle: Diffraction effects are observed for beams of electrons scattered by the crystals.
λ = \(\frac { h }{ p } =\frac { h }{ \sqrt { 2mE_{ k } } } =\frac { h }{ \sqrt { 2meV } } \)
= \(\frac { 6.63\times { 10 }^{ -34 } }{ \sqrt { 2\times 9.1\times 10^{ -31 }\times 1.6\times 10^{ -19 }\times 120 } } \)
λ = 0.112 nm.

(i) λ_e = \(\frac { h }{ { p }_{ e } } \) and λ_p = \(\frac { h }{ { p }_{ e } } \) , λ_e-λ_p = 1.00 nm
So, \(\frac { \lambda _{ e } }{ { \lambda }_{ p } } =\frac { { p }_{ e } }{ { p }_{ e } } =\frac { 1 }{ 1 } \Rightarrow \frac { { p }_{ p } }{ { p }_{ e } } =\frac { 1 }{ 1 } \) = 1:1
(ii) From relation K = \(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { { p }^{ 2 } }{ 2m } \)
K_e = \(\frac { { p }_{ e }^{ 2 } }{ 2m_{ e } } \) and K_p = \(\frac { { p }_{ p }^{ 2 } }{ 2m_{ p } } \)
\(\frac { { K }_{ p } }{ { K }_{ e } } =\frac { { p }_{ p }^{ 2 } }{ { 2m }_{ p } } \times \frac { { 2m }_{ e } }{ { p }_{ e }^{ 2 } } =\frac { m_{ e } }{ { m }_{ p } } \)
Since m_e <<< m_p. So K_p <<< K_e
\(\frac { { K }_{ p } }{ { K }_{ e } } =\frac { 9.1\times 10^{ -31 } }{ 1.67\times 10^{ -27 } } \) = 5.4 x 10^-4.

In ground state, the kinetic energy of the electron is
K = -E =  \(\frac { +13.6eV }{ 1^{ 2 } } \) =13.6 x 1.6 x 10^-19 J
de Broglie wavelength,
λ  = \(\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } } \)
λ₁ = -\(\frac { 6.63\times { 10 }^{ -34 } }{ \sqrt { 2\times 9.1\times 10^{ -31 }\times 2.18\times 10^{ -18 } } } \)
= 9.33 x 10^-9 = 0.33 nm
Kinetic energy in the first excited state (n = 2)
K = -E = + \(\frac { 13.6 }{ 2^{ 2 } } \) eV = +3.4 eV
= 3.4 x 1.6 x 10^-19 J
= 0.54 x 10^-18 J
de Broglie wavelength,
λ₂ = \(\frac { h }{ \sqrt { 2mK } } \)
=  \(\frac { 6.63\times { 10 }^{ -34 } }{ \sqrt { 2\times 9.1\times { 10 }^{ -31 }\times 0.544\times { 10 }^{ -15 } } } \)
= 2 x 0.33 nm = 0.66 mm.

Given data: P = 5 x 10^-3 W,
n = 8 x 10¹⁵ photons per second
Energy of each photon,
E = \(\frac { p }{ n } =\frac { 5\times { 10 }^{ -3 } }{ 8\times { 10 }^{ 15 } } \)
= 6.25 x 10^-19 J
= \(\frac { 6.25\times 10^{ -19 }J }{ 1.6\times { 10 }^{ -19 } } \)
E = 3.9 eV
Work function, W_o = E-V_o
= (3.9-2)eV = 1.9 eV

(i) The threshold frequency is the frequency of incident light at which kinetic energy of ejected photoelectron is zero.
∴ From fig. threshold frequency,
v₀ = 4.5 x 10¹⁴ Hz
(ii) Work function, W = hv₀
= 6.6 x 10^-34 x 4.5 x 10¹⁴ joule
= \(\frac { 6.6\times { 10 }^{ -34 }\times 4.5\times 10^{ 14 } }{ 1.6\times { 10 }^{ -19 } } \) eV
= 1.85 eV

(i) de Broglie wavelength associated with a moving charge particle having a KE 'K' can be given as
λ = \(\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } } \)      ......(1)  \(\left[ K=\frac { 1 }{ 2 } mv^{ 2 }=\frac { { p }^{ 2 } }{ 2m } \right] \)
(ii) The kinetic energy of the electron in any orbit of the hydrogen atom can be given as
K = -E = -\(\left( \frac { 13.6 }{ { n }^{ 2 } } eV \right) =\frac { 13.6 }{ { n }^{ 2 } } \)       ........(2)
(iii) Let K and K₄ be the KE of the electron in the ground state and third excited state, where n₁ = 1 shows the ground state and n₂ = 4 shows the third excited state.
Using the concept of equations (1) & (2), we have
\(\frac { { \lambda }_{ 1 } }{ { \lambda }_{ 4 } } =\sqrt { \frac { { K }_{ 4 } }{ { K }_{ 1 } } } =\sqrt { \frac { { n }_{ 1 }^{ 2 } }{ { n }_{ 2 }^{ 2 } } } \)
\(\frac { { \lambda }_{ 1 } }{ { \lambda }_{ 4 } } =\sqrt { \frac { { 1 }^{ 2 } }{ { 4 }^{ 2 } } } =\frac { 1 }{ 4 } \)
λ = \(\frac { { \lambda }_{ 4 } }{ 4 } \)
i.e., the wavelength in the ground state will decrease.

X-rays are being used in many fields. Let us list a few of them.
(i) Medical diagnosis X-rays can pass through flesh more easily than through bones. Thus an x-ray radiograph containing a deep shadow of the bones and a light shadow of the flesh may be obtained. X-ray radiographs are used to detect fractures, foreign bodies, diseased organs, etc.
(ii) Medical therapy Since x-rays can kill diseased tissues, they are employed to cure skin diseases, malignant tumors, etc.
(iii) Industry X-rays are used to check for flaws in welded joints, motor tires, tennis balls, and wood. At the customs post, they are used for the detection of contraband goods.
(iv) Scientific research X-ray diffraction is an important tool to study the structure of the crystalline materials - that is, the arrangement of atoms and molecules in crystals.

(i) X-ray spectra show some narrow peaks at some well-defined wavelengths when the target is hit by fast electrons.
(ii) The line spectrum showing these peaks is called the characteristic x-ray spectrum. This x-ray spectrum is due to the electronic transitions within the atoms.
(iii) When an energetic electron penetrates into the target atom and removes some of the K-shell electrons.
(iv) Then the electrons from outer orbits jump to fill up the vacancy so created in the K-shell.
(v) During the downward transition, the energy difference between the levels is given out in the form of an x-ray photon of a definite wavelength.

(vi) Such wavelengths, characteristic of the target, constitute the line spectrum.
(vii) It is evident that K-series of lines in the x-ray spectrum of an element arises due to the electronic transitions from L, M, N, levels to the K-level. Similarly, the longer wavelength L-series originates when an L-electron is knocked out of the atom and the corresponding vacancy is filled by the electronic transitions from M, N, 0, ..., and so on.
(viii) The K\(\alpha\) and K\(\beta\) of the K-series of molybdenum are shown by the two peaks in its x-ray spectrum.

Energy differnce = Energy of emitted photon
= E₂ - E₁
= -1.51 - (-3.4) = 1.89 eV
= 1.89 x 1.6 x 10^-19 J
λ = \(\\ \frac { hc }{ { E }_{ 2 }-{ E }_{ 1 } } \)
= \(\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 1.89\times 1.6\times 10^{ -19 } } =\frac { 19.8 }{ 3.024 } \) x 10^-7
= 6.548 x 10^-7 m = 6548 Å
The wavelength belongs to the Balmer series of the hydrogen spectrum.

The energy of electrons in the nth orbit of a hydrogen atom is
En = -\(\frac { 13.6 }{ { n }^{ 2 } } \)
when the incident beam of energy 12.3 eV is absorbed by a hydrogen atom. Let the electron jump from n = 1 to n = n level.
E = E_n - E₁
12.3 = \(\frac { 13.6 }{ { n }^{ 2 } } -\left( -\frac { 13.6 }{ { I }^{ 2 } } \right) \)
⇒ 12.3 = 13.6 \(\left| 1-\frac { 1 }{ n^{ 2 } } \right| \Rightarrow \frac { 12.3 }{ 13.6 } =1-\frac { 1 }{ { n }^{ 2 } } \)
⇒ 0.9 = \(1-\frac { 1 }{ { n }^{ 2 } } \)  ⇒ n² = 10 ⇒ n = 3
That is the hydrogen atom would be excited up to the second excited state.
For Lyman Series
\(\frac { 1 }{ \lambda } =R\left[ \frac { 1 }{ { n }_{ f }^{ 2 } } -\frac { 1 }{ { n }_{ i }^{ 2 } } \right] \)
⇒ \(\frac { 1 }{ \lambda } \) =0.097 x 10⁷ \(\left[ \frac { 1 }{ 1 } -\frac { 1 }{ 9 } \right] \)
⇒ \(\frac { 1 }{ \lambda } \) =1.097 x 10⁷ x \(\frac { 8 }{ 9 } \)
λ =\(\frac { 9 }{ 8\times 1.097\times { 10 }^{ 7 } } \)
=1.025 x 10⁷ =102.5 mm
For Balmer Series
\(\frac { 1 }{ \lambda } \) = 1.097 x 10⁷ \(\left[ \frac { 1 }{ 4 } -\frac { 1 }{ 16 } \right] \)
⇒ \(\frac { 1 }{ \lambda } \) = 1.097 x 10⁷ x \(\frac { 3 }{ 16 } \)
⇒ λ = 4.86 x 10^-7 m ⇒ λ = 486 nm

Let r be the center to center distance between the alpha particle and the nucleus (Z = 80). When the alpha particle is at the stopping point, then
K = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { (Ze)(2e) }{ r } \)
(or) r = \(\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } .\frac { 2Ze^{ 2 } }{ K } \)
= \(\frac { 9\times { 10 }^{ 9 }\times 2\times 80{ e }^{ 2 } }{ 4.5MeV } \)
= \(\frac { 9\times { 10 }^{ 9 }\times 2\times 80\times (1.6\times { 10 }^{ -19 })^{ 2 } }{ 4.5\times { 10 }^{ 6 }\times 1.6\times { 10 }^{ -19 }J } \)
= \(\frac { 9\times 160\times 16 }{ 4.5 } \) x 10^-16 = 512 x 10^-16 m
= 5.12 x 10^-14 m

i) When a fast-moving electron penetrates and approaches a target nucleus, the interaction between the electron and the nucleus either accelerates or decelerates it which results in a change of path of the electron.
ii) The radiation produced from such decelerating electrons is called Bremsstrahlung or braking radiation.

iii) The energy of the photon emitted is equal to the loss of kinetic energy of the electron
iv) Since an electron may lose part or all of its energy to the photon, the photons are emitted with all possible energies (or frequencies).
v) The continuous x-ray spectrum is due to such radiations
vi) When an electron gives up all its energy, then the photon is emitted with the highest frequency v_o (or lowest wavelength \(\lambda\)_o).
vii) The initial kinetic energy of an electron is given by eV where V is the accelerating voltage. Therefore, we have
\({ hv }_{ 0 }=eV(or)\frac { hc }{ { \lambda }_{ 0 } } =eV\)
\(\\ { \lambda }_{ 0 }=\frac { hc }{ eV } \)
viii) Where \({ \lambda }_{ 0 }\) is the cut-off wavelength. Substituting the known values in the above equation, we get
\({ \lambda }_{ 0 }=\frac { 12400 }{ V } \)Å    ...(1)
The relation given by equation (1) is known as the Duane - Hunt formula.
ix) The value  \({ \lambda }_{ 0 }\)depends only on the accelerating potential and is the same for all targets.
x) This is in good agreement with the experimental results.
xi) Thus, the production of continuous x-ray spectrum and the origin of cut-off wavelength can be explained on the basis of the photon theory of radiation.

i) X-rays are produced in x-ray tube which is essentially a discharge tube.
ii) A tungsten filament F is heated to incandescence by a battery. As a result, electrons are emitted from it by thermionic emission.

iii) The electrons are accelerated to high speeds by the voltage applied between the filament F and the anode.
iv) The target materials like tungsten, molybdenum are embedded in the face of the solid copper anode.
v) The face of the target is inclined at an angle with respect to the electron beam so that x-rays can leave the tube through its side.
vi) When high-speed electrons strike the target, they are decelerated suddenly and lose their kinetic energy.
vii) As a result, x-ray photons are produced. Since most of the kinetic energy of the bombarding electrons gets converted into heat, targets' made of high-melting-point metals and a cooling system are usually employed.