Dual Nature of Radiation and Matter

In solar cells, electron emission converts the light energy into electrical energy through the photoelectric effect. Solar cells work due to photoelectric effect, in which photons from sunlight strike solar cell surface. When a photon that has sufficient energy hits the surface, it transfers its energy to an electron in the material of solar cell surface. This causes the electron emission from its bound state. Solar cells can use photoelectric effect to generate electron-hole pairs that are separated and collected for producing electrical current. This provides a renewable source of power.

Electron emission is used for generating a beam of electrons in an electron microscope. An electron microscope requires a source of electrons for creating a beam of electrons. This source can be mostly a gun that uses electron emission for producing free electrons. Once the electrons are emitted, they are accelerated by anode and then, they are focused as a fine beam through a series of electromagnetic lenses. The beam is then directed onto the specimen that is being examined.

The work function in electron emission is the minimum amount of energy needed for removing an electron from the surface of a material. This function is denoted as? , and it is measured in electron volts. An electron requires at least the work function as energy to escape the surface of a solid material into the vacuum. This represents the difference between energy of an electron at rest in vacuum and fermi level of the material. Work function varies from material to material.

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt[]{2km}}=\frac{h}{2\left(ev\right)m}$

$\lambda \alpha \frac{1}{\sqrt[]{m}}\Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=\sqrt[]{\frac{m_p}{m_e}}$

Let the work function of metal is $?$ , so

$k_1=3V_{0}e=\frac{hc}{\lambda }-?$ . (i)

$K_2=V_{0}e=\frac{hc}{2\lambda }-?$ . (ii)

Doing (i) – 3 X (ii), we have

$0=\frac{hc}{\lambda }-\frac{3hc}{2\lambda }+2?\Rightarrow ?=\frac{hc}{4\lambda }$    

The threshold wavelength = $\frac{hc}{?}=4\lambda$

$h\nu =3.4-1.51=1.89eV$

As we know that radius of circular path in magnetic field is given as

$r=\frac{\sqrt[]{2mK}}{qB}\Rightarrow K=\frac{r^2{q}^2{B}^2}{2m}$              

$\Rightarrow K=\frac{{\left(7*10^{-3}\right)}^2*\left(1.6*10^{-19}\right)*{\left(5*10^{-4}\right)}^2}{2*9.1*10^{-31}}eV=107.7*10^{-2}eV$  = 1.08 eV

$\Rightarrow ?=h\nu -K=1.89-1.08=0.81eV$              

According to Doppler's effect in light, we can write

$\frac{v}{c}=\frac{\Delta \lambda }{\lambda }\Rightarrow \frac{v}{c}=\frac{5896-5890}{5890}=\frac{6}{5890}=\frac{3}{2945}\Rightarrow v=\frac{3}{2945}*3*10^5\approx 305.6km/s$

Schrödinger's wave equation explains the microscopic world that governs chemistry, electronics, and modern technology. It's basically how our digital world actually functions at the atomic level.

Computer chips, lasers, MRI machines, LED lights, and solar panels all rely on quantum effects that this equation of Schrodinger helps us understand and design.

The Schrodinger wave equation is the foundation of quantum mechanics. Without it, we couldn't understand how atoms work or predict the behaviour of electrons in materials.