Dual Nature of Radiation and Matter

11.7 Power of the sodium lamp, P = 100 W

Wavelength of the emitted sodium light, $\lambda$ = 589 nm = 589 $*{10}^{-9}$ m

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

Speed of light, c = 3 $*{10}^8$ m/s

Energy per photon associated with the sodium light is given as:

$E$ = $\frac{hc}{\lambda }$ = $\frac{6.626\mathrm{}*{10}^{-34}*3\mathrm{}*{10}^8}{589\mathrm{}*{10}^{-9}}$ = 3.37 $*{10}^{-19}$ J = $\frac{3.37\mathrm{}*{10}^{-19}}{1.6*{10}^{-19}}$ eV = 2.11 eV

Let the number of photon delivered to the sphere = n

The equation of power can be written as $P=nE$

$n$ = $\frac{P}{E}$ = $\frac{100}{3.37\mathrm{}*{10}^{-19}}$ photons/sec = 2.97 $*{10}^{20}$ photons/s

Therefore, every second, 2.97 $*{10}^{20}$ are delivered to the sphere.

Therefore, every second, 2.97 $*{10}^{20}$ are delivered to the sphere.

11. The slope of the cut-off voltage (V) versus frequency ( $\nu )$ of an incident light is given as:

$\frac{\mathrm{V}}{\nu }=$ 4.12 $*{10}^{-15}$ Vs

The relationship of V and $\nu$ is given as $h\nu$ = $eV$ or $\frac{\mathrm{V}}{\nu }=$ $\frac{h}{e}$

where e = Charge of an electron = 1.6 $*{10}^{-19}$ C and h = Plank's constant

Therefore, h = $e*\frac{\mathrm{V}}{\nu }$ = 1.6 $*{10}^{-19}$ $*$ 4.12 $*{10}^{-15}$ = 6.592 $*{10}^{-34}$ Js

Plank's constant = 6.592 $*{10}^{-34}$ Js

11.5 The energy flux of sunlight, $?$ = 1.388 $*{10}^3$ W/ $m^2$

Hence power of the sunlight per square meter, P = 1.388 $*{10}^{3}W$

Speed of light, c = 3 $*{10}^8$ m/s

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

Average wavelength of photon, $\lambda$ = 550 nm = 550 $*{10}^{-9}$ m

If n is the number of photon per square meter, incident on earth per second, the equation of power can be written as

$P=nE$ or $n$ = $\frac{P}{E}$

We know $E$ = $\frac{hc}{\lambda }$

Hence, $n$ = $\frac{P\lambda }{hc}$ = $\frac{1.388*{10}^3*550\mathrm{}*{10}^{-9}}{6.626\mathrm{}*{10}^{-34}*3\mathrm{}*{10}^8}$ = 3.84 $*{10}^{21}$ photons $m^2$ /s

Therefore, every second 3.84 $*{10}^{21}$ photons are incident per square meter on earth.

11.3 Photoelectric cut-off voltage,  $V_0$ = 1.5 V

Maximum kinetic energy of the photoelectrons emitted is given as

$K=e{V}_0$ , where e = charge of an electron = 1.6 $*{10}^{-19}$ C

K = 1.6 $*{10}^{-19}*1.5$ = 2.4 $*{10}^{-19}$ J

Hence, maximum kinetic energy of the photoelectrons emitted is 2.4 $*{10}^{-19}$.

11.1 Potential of the electrons, V = 30 kV = 3 $*{10}^4$  V

Energy of the electron, E = e $*V$  where e = charge of an electron = 1.6 $*{10}^{-19}C$

(a) Maximum frequency produced by the X-ray = $\nu$

The energy of the electron is given by the relation, E = h $\nu$ ,

where h = Planck's constant = 6.626 $*{10}^{-34}$  Js

$\nu =\frac{E}{h}$ = $\frac{3\mathrm{}*{10}^4*1.6*{10}^{-19}}{6.626*{10}^{-34}}$  = 7.244 $*{10}^{18}$  Hz

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{u}\mathrm{m}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{X}-\mathrm{r}\mathrm{a}\mathrm{y}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}$ 7.244 $*{10}^{18}$  Hz

(b) The minimum wavelength produced is given as

  $\lambda =\frac{c}{\nu }$ where c = Speed of light in air, c = 3 $*{10}^{18}$  m/s

  $\lambda =\frac{3\mathrm{}*{10}^8}{7.244*{10}^{18}}$ = 4.14 $*{10}^{-11}$  m = 0.0414 nm

Hence, the minimum wavelength produced is 0.414 nm.

The Davisson-Germer experiment was conducted to test the De-broglie hypothesis and know wave nature of electrons. In this experiment, a beam of electrons was emitted from an electron gun with known acceleration to strike a nickel crystal placed inside a vacuum chamber.

After striking the surface, electrons are scattered from the crystal surface at different angles. The diffraction pattern obtained through this experiment was similar to that produced by X-rays, which is wave.

The Davisson-Germer experiment provided the first experimental proof of the wave nature of matter and de Broglie's hypothesis right.

Significance of de Broglie's Hypothesis in Modern Physics (NCERT-based answer):

De Broglie's hypothesis is a revolutionary idea which changed the understanding of matter and wave as we know it. This hypothesis introduced the concept that particles of matter, like electrons, and Proton, can behave like wave and have wave-like properties, just as light exhibits both wave and particle nature.

According to de Broglie, any moving particle has a wavelength given by:

$\lambda = \frac {h} {p} = \frac {h} {mv}$

where $\lambda$ is the wavelength, $h$ is Planck's constant, $m$ is the mass, and $v$ is the velocity of the particle.

The photoelectric effect when light is projected in a metal surface, it's surface ejects electron (which are in outermost shell or free electron). This happens because the light provides additional energy to the electron to detach itself from metal surface. Well to do this light must have a certain amount of energy which in other words can be said the right frequency of light is required for photoelectric effect. 

This frequency must be above a certain minimum value (called threshold frequency) fixed for every metal, regardless of its intensity. Photoelectric effect cannot be explained by the wave theory of light.