Quantum Theory of Light: Wave-Particle Duality, Planck’s Constant, Corpuscular Theory

Ans. The wavelength and energy are related by the formula ΔE=hc / λ,

where h is Planck's constant (= 6.626 x 10^-34 Js), c is the speed of light (3.00 x 10⁸ m/s) and λ is the wavelength in meters. 

The wavelength can then be calculated by rearranging the above formula as follows: 

λ = hc / E (Energy of Photon)

= 3.79 x 10^-7 Joules

Ans. We know, E=E₀+1/2 mv²

We know, E=hc / λ₀

=(6.626 x 10^-34 J/s) x  

= 3.31 x 10^-19J

E=hc / λ

=(6.626 x 10^-34 J/s) x  

= 4.97 x 10^-19J

Or E= 1.66 x 10^-19J

Ans. Light exhibits the behavior of both a particle and a wave. The Transactional Interpretation (TI) shows us that light moves from place to place as waves, but at locations where it is emitted or absorbed it obeys the boundary conditions of a particle carrying energy ω, momentum k, and angular momentum. 

One might say that light travels as a wave but takes off and lands as a particle. It is usually convenient to think of low energy photons, e.g., radio transmissions, as waves, and high energy photons like X-rays and gamma rays as particles. In the TI the wavelike behavior is present in the offer and confirmation waves, and the particle-like behavior is present in the completed transactions. 

Ans. Setting to zero is a way of taking the classical limit of quantum mechanics: the uncertainty principle goes away, and in that limit, quantum mechanics becomes more or less equivalent to Newtonian mechanics. 

However, Planck’s treatment of black body radiation suggests that a universe in which Planck’s constant was zero would be very different than ours because hot objects would rapidly radiate away all their energy as high-frequency electromagnetic radiation (the ultraviolet catastrophe), matter would be extremely cold, and the universe would be dominated by light.

Ans. Light waves have a characteristic frequency f indicating how many times per second the electric field of the light wave oscillates. The angular frequency ω is just the same characteristic expressed in radians of phase per second instead of oscillations per second, so ω = 2π f. 

Light waves also have a characteristic wavelength λ as they move through space, which is the spatial distance between one electric-field maximum and the next. The wavenumber k is a way of looking at the reciprocal of that characteristic, so that k = 2π/λ. 

The speed of light c is related to these quantities as c = f λ = ω/k. When we deal with particle waves, for example using the de Broglie wavelength, we also characterize them in terms of ω and k.

Ans. The Quantum Theory of Radiations by Max Planck can explain the photoelectric effect very well. Max Planck says that light is made of bundles of energy known as photons, energy of each photon being equal to hv,v being the frequency of the light. 

So, when a photon of frequency being equal to the threshold frequency of the metal strikes the metal surface, it imparts its whole energy to an electron and hence helps the electron to overcome the forces of attraction from the nucleus of the atom. In this way, an electron is ejected out of a metal surface when a photon strikes it. 

If the frequency of the photon is less than v₀ (threshold frequency), then no photoelectric effect is observed.

But if v>v_o then some of its energy (which is equal to the ionization energy) is consumed in ejecting the electron from the metal and the remaining is used to impart some kinetic energy to the ejected electron.

hv=Φ+ 12mv²

Here, Φ is the ionization energy, v is the velocity imparted to the ejected electron, m is the mass of the electron.

As Φ=hv_o

∴ hv=1/2mv²+hvo

Ans. A photon is an elementary particle that is a carrier of electromagnetic fields or radiation. They are massless particles in the sense that their mass is zero at zero momentum. They always move at the speed of light c in free space and are the basic units of all mass. This property is also known as the invariant mass and remains independent of overall momentum.

We are given that the energy possessed by a photon is given as E=hν.

But from the mass-energy equivalence principle which states that a body in motion possesses relativistic mass that can be derived from its total energy divided by the speed of light squared, we see that energy possessed by a photon in motion can be quantified as:

E=mc²

Equating the two energy equations we get:

⇒hν=mc²⇒m=hv/c²

Let the photons fall through a distance s when a horizontal flash of light is sent through the earth’s gravitational field. The photons follow the kinematics of a projectile motion.

The photons travel with a horizontal velocity of c but have no vertical velocity associated with their motion. 

Therefore, u_x=c and u_y=0

If the horizontal distance travelled by a photon following this trajectory is d_x=d, then the time it takes to travel this distance is given by:

t=d_x/u_x=dc

Therefore, the distance through which the photon would fall vertically downwards in this time can be given by the kinematic equation of motion:

h_y=u_yt+1/2gt²

⇒s=0+1/2gt²

s=1/2g.(d/c)²=gd² / 2c²

Therefore, the correct option would be A