Modern Physics Concept Page - 5

Diagram

Interception of the plot

• Slope: From the above graph we can see that the slope of the graph is constant,i.e., he${\displaystyle \frac{h}{e}}$. 
• Intercept: The intercept is −We${\displaystyle \frac{-W}{e}}$. This is negative y intercept. This intercept depends upon work function of the metal, W$W$. More the work function, more will be the magnitude of the intercept. 

Law

Max Planck's Quantum Theory

When black body is heated, it emits thermal radiations of different wavelengths or frequency. To explain these radiations, max planck put forward a theory known as plancks quantum theory.
(i) The radiant energy which is emitted or absorbed by the black body is not continuous but discontinuous in the form of small discrete packets of energy, each such packet of energy is called a 'quantum'. In case of light, the quantum of energy is called a 'photon'.
(ii) The energy of each quantum is directly proportional to the frequency (v)$(v)$ of the radiation, i.e. E∝V$E\propto V$ or E=hv=hc/$E=hv=hc/$ Where, $$$$ Planck's constant $$= 6.62\times 10^{27} erg. \ sec.$$ or $$6.6210^{-24} joules \ sec$$
 (iii) The total amount of energy emitted or absorbed by a body will be some whole number quanta. Hence $$E = nhv$$ where $$n$$ is an integer.

Example

Planck's constant using de-broglie equation

If U.V. light of wavelengths 800 A0$A^0$ and 700 A0$A^0$ can liberate electrons with kinetic energies of 1.8 eV$1.8eV$ and 4 eV$4eV$ respectively from hydrogen atom in ground state, then the value of Planck's constant is found as follows :From first eqn of photoelectric effect,
⇒(hc8×108)−W=1.8$\Rightarrow ({\displaystyle \frac{hc}{8}}\times {10}^8)-W=1.8$ ---------------(I)

From second equation of photoelectric effect,
⇒hc×1087−W=4$\Rightarrow {\displaystyle \frac{hc\times {10}^8}{7}}-W=4$ -----------------(II)

Solving (I) and (II) simultaneously we get,
h=4.10625×10−15 eV$h=4.10625\times {10}^{-15}eV$

Converting this h$h$ in terms of Js$Js$ we get,
h=4.10625×1.6×10−19−15$h=4.10625\times 1.6\times {10}^{-19-15}$
h=6.57×10−34 Js$h=6.57\times {10}^{-34}Js$

Example

Einstein's photoelectric equation example

Three metals have work functions in the ratio 2:3:4.Graphs are drawn for all between the stopping potential and the incident frequency. The graphs have slopes in the ratio:Here equation is
V=(hν−ϕ)/e$V=(h\nu -\varphi )/e$
or V=hνe−ϕe$V={\displaystyle \frac{h\nu }{e}}-{\displaystyle \frac{\varphi }{e}}$
or comparing with
y=mx+c$y=mx+c$
we get m=h/ec=−ϕ/e$m=h/ec=-\varphi /e$
So, slope (m)=h/e$(m)=h/e$
which is a constant.
So, ratio is 1:1:1.

Example

Photocurrent for a given intensity

Let the intensity of the incident photons be I$I$ and frequency be ν$\nu$.
Then, the incident power is P=IA$P=IA$ where A$A$ is the area of the conductor.
Energy of one photon, Ep=hcλ$E_p={\displaystyle \frac{hc}{\lambda }}$
Then, number of photons incident per unit time is given by, n=PEp$n={\displaystyle \frac{P}{E_p}}$
Since, each photon emits one electron, number of electrons emitted is n$n$.
Photocurrent is given by Ip=ne$I_p=ne$ where e:$e:$ charge of an electron.

Definition

De-broglie wavelength

De Broglie's wavelength is the wavelength associated with a massive particle,hypothesized by De Broglie that explains Bohr's quantised orbits by bringing in the wave-particle duality.  It is written as
λ=hmv$\lambda ={\displaystyle \frac{h}{mv}}$ (de broglie wavelength)

Example

Mass of a particle using de-Broglie wavelength

The de-Broglie wavelength associated with a particle moving with a velocity of 102${10}^2$ m/s is 6.6×10−24m$6.6\times {10}^{-24}m$.Then the mass of the particle is 
λ=hmv$\lambda ={\displaystyle \frac{h}{mv}}$⇒m=hυλ=6.6×10−346.6×10−24×102=10−12kg$\Rightarrow m={\displaystyle \frac{h}{\upsilon \lambda }}={\displaystyle \frac{6.6\times {10}^{-34}}{6.6\times {10}^{-24}\times {10}^2}}={10}^{-12}kg$

Definition

Heisenberg uncertainty principle

The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision.
ΔxΔp>ℏ2$\mathrm{\Delta }x\mathrm{\Delta }p>{\displaystyle \frac{\hslash }{2}}$ΔEΔt>ℏ2$\mathrm{\Delta }E\mathrm{\Delta }t>{\displaystyle \frac{\hslash }{2}}$

Formula

Uncertainty in position or momentum

ΔxΔp>ℏ2$\mathrm{\Delta }x\mathrm{\Delta }p>{\displaystyle \frac{\hslash }{2}}$
where Δx$\mathrm{\Delta }x$ is the uncertainty in position,
Δp$\mathrm{\Delta }p$ is the uncertainty in momentum and
ℏ$\hslash$ is the reduced Planck's constant

Example

Calculate the de broglie wavelength of an electron accelerated by a given potential

Example: An electron of charge e and mass m is accelerated from rest by a potential difference V. Find the de-Broglie wavelength.

Solution:
12mv2=eV$\frac{1}{2}m{v}^2=eV$
mv=2eVm−−−−−√$mv=\sqrt{2eVm}$
So de-broglie wavelength
λ=hmv$\lambda ={\displaystyle \frac{h}{mv}}$
=h2meV−−−−−√$={\displaystyle \frac{h}{\sqrt{2meV}}}$

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