\(\lambda_0  =\frac{h c}{\phi}  \)
\(=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.313 \times 1.6 \times 10^{-19}}  \)
\( =\frac{19.8782400}{5.3} \times 10^{-7}  \)
\(\lambda_0  =3.750 \times 10^{-7} \simeq 3750 \stackrel{o}A\)

\(\lambda\propto \frac{1}{\sqrt{V}}\)
\(\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{224\times10^3}}{\sqrt{14\times 10^3}}\)
\(=\sqrt{16}=4\)
\(\lambda_{\mathrm{2}}= \frac{\lambda_1}{4}\)

No, most electrons get scattered into the metal. Only a few come out of the surface of the metal.

The minimum retarding (negative) potential of anode of a photoelectric tube fro which photoelectric current stops or becomes zero is called the stopping potential.

de Broglie wavelength \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\)
Kinetic energy of electron K \(=\frac{1}{2} m v^{2} \text { (or) } v=\sqrt{\frac{2 K}{m}}\)
Now de Broglie wavelength is \(\lambda=\frac{h}{m v}=\frac{h}{m \sqrt{2 K / m}} \)
\(\lambda=\frac{h}{\sqrt{2 m K}} \)

The de Broglie wavelength associated with the kinetic energy K is \(\lambda=\frac{h}{\sqrt{2 m k}}\)
Where m is the mass of the particle
Since proton and electron have same KE, the wavelength is inversely proportional to square root of the mass \(\lambda \alpha \frac{1}{\sqrt{m}}\)
Mass of proton is 1840 times greater them that of electron. Therefore, de Broglie's wavelength of electron is greater than the proton.

According to quantum concept, intensity of light of given wavelength is defined as the number of energy quanta or photons incident per unit area per unit time, with each photon having same energy. Its unit is Wm^-2.

(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.

Principle:
(i) The wave nature of the electron is used in the construction of microscope called electron microscope.
(ii)The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under study.
(iii) Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths.
(iv) De Broglie's wavelength of electron is very much less than (a few thousand less) that of the visible light being used in optical microscopes.
(v) As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope.
(vi)  Electron microscopes giving magnification more than 2,00,000 times are common in research laboratories.

Working:
(i) The construction and working of an electron microscope is similar to that of an optical microscope except that in electron microscope focussing of electron beam is done by the electrostatic or magnetic lenses.
(ii) The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done.
(iii) The electrons emitted from the source are accelerated by high potentials.
(iv) The beam is made parallel by magnetic condenser lens; When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample.
(v) With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all brands of science.

Explanation for the photoelectric effect:
The experimentally observed facts of photoelectric effect can be explained with the help of Einstein's photoelectric equation.
(i) As each incident photon liberates one electron, then the increase of intensity of the light (the number of photons per unit area per unit time) increases the number of electrons emitted thereby increasing the photocurrent. The same has been experimentally observed.
(ii) From K_max = hv - Φ_0, it is evident that K_max is proportional to the frequency of the light and is independent of intensity of the light.
(iii) As given in equation \({ hv }_{ o }+\cfrac { 1 }{ 2 } { mv }^{ 2 }\) , there must be minimum energy (equal to the work function of the metal) for incident photons to liberate electrons from the metal surface. Below which, emission of electrons is not possible. Correspondingly, there exists minimum frequency called threshold frequency below which there is no photoelectric emission.
(iv) According to quantum concept, the transfer of photon energy to the electrons is instantaneous so that there is no time lag between incidence of photons and ejection of electrons.
Thus, the photoelectric effect is explained on the basis of quantum concept of light.

(i) The photoemission of electrons does not depend on the intensity but it depends on the frequency and hence on the energy of a photon of incident light.
(ii) If the energy of a photon is greater than the work function, the photoemission of electrons results however weak the incident radiation may be.
(iii) The energy of a photon of red light is less than the work function of zinc, so red light cannot emit photoelectrons.
(iv) The energy of a photon of ultraviolet light is greater than the work function of zinc, so ultraviolet light can emit photoelectrons.

Curves a and b have different intensities but the same stopping potential, so curves 'a' and 'b' have the same frequency but different intensities.

i) The momentum of photon of frequency v is given by,
\(p=\frac { hv }{ c } =\frac { h }{ \lambda } \)
ii) The wavelength of a photon in terms of its momentum is,
\(\lambda =\frac { h }{ p } \)
iii) According to de Broglie, the above equation is completely a general one and this is applicable to material particles as well. Therefore, for a particle of mass m traveling with speed u, the wavelength is given by,
\(\lambda =\frac { h }{ mv } =\frac { h }{ p } \)
iv) This wavelength of the matter waves is known as de Broglie wavelength. This equation relates the wave character (the wavelength \(\lambda\)) and the particle character (the momentum p) through Planck's constant.

\(v= 9 \times 10^{14} \mathrm{~Hz} ; \mathrm{V}=8 \times 10^{5} \mathrm{~m} / \mathrm{s} \)
\(\mathrm{E}= \mathrm{h}-\mathrm{h} v_{o} \Rightarrow v_{o}=\frac{\mathrm{h}v-\mathrm{E}}{\mathrm{h}}=\mathrm{V}-\frac{\mathrm{E}}{\mathrm{h}}=\mathrm{V}-\frac{\frac{1}{2} \mathrm{mv}^{2} }{h}\)
\(v_s={9 \times 10^{14}-\frac{\left[\frac{1}{2} \times 9.1 \times 10^{-31} \times\left(8 \times 10^{5}\right)^{2}\right]}{6.626 \times 10^{-34}}}=4.605 \times 10^{14} \)
\( v_{o} \simeq 4.6 \times 10^{14} \mathrm{~Hz} \)

\(\lambda=4000 \stackrel{o}A=4000 \times 10^{-10} \mathrm{~m} \)
\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\Rightarrow \mathrm{v }=\frac{\mathrm{h}}{\mathrm{m\lambda}}=\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 4000 \times 10^{10}}=1820 \mathrm{~ms}^{-1} \)