Dual Nature of Radiation and Matter

11.13 Kinetic energy of electron, $E_k=$ 120 eV = 120 $*1.6*{10}^{-19}$ J

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

Charge of an electron, e = 1.6 $*{10}^{-19}$ C

Mass of electron, m = 9.1 $*{10}^{-31}$ kg

The kinetic energy of electron can be written as $E_k=\frac{1}{2}$ m $v^2$

$v$ = $\sqrt{\frac{2E_k}{m}}$ = $\sqrt{\frac{2*120\mathrm{}*1.6*{10}^{-19}}{9.1\mathrm{}*{10}^{-31}}}$ = 6.496 $*{10}^6$ m/s

Momentum of the electron, p = mv = 9.1 $*{10}^{-31}*$ 6.496 $*{10}^6$ = 5.911 $*{10}^{-24}$ kgm/s

Speed of the electron = 6.496 $*{10}^6$ m/s

De Broglie wavelength of an electron with momentum p is given as

$\lambda =\frac{h}{p}$ = $\frac{6.626\mathrm{}*{10}^{-34}}{5.911\mathrm{}*{10}^{-24}}$ = 1.121 $*{10}^{-10}$ m = 0.1121 nm

11.12 Potential difference, V = 56V

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

Charge of an electron, e = 1.6 $*{10}^{-19}$ C

Mass of electron, m = 9.1 $*{10}^{-31}$ kg

At equilibrium, the kinetic energy of each electron is equal to the acceleration potential. If v is the velocity of each electron, we can write

$\frac{1}{2}m{v}^2$ = $eV$

$v=\sqrt{\frac{2eV}{m}}$ = $\sqrt{\frac{2*1.6*{10}^{-19}*56}{9.1\mathrm{}*{10}^{-31}}}$ = 4.44 $*{10}^6$ m/s

The momentum of each electron = mv = 9.1 $*{10}^{-31}*$ 4.44 $*{10}^6$ kgm/s = 4.04 $*{10}^{-24}$ kgm/s

De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:

$\lambda =\frac{12.27}{\sqrt{V}}$ Å = $\frac{12.27}{\sqrt{56}}$ Å = 1.64 Å = 1.64 $*{10}^{-10}$ m = 0.164 nm

11.11 Wavelength of the light produced by Argon laser, $\lambda$ = 488 nm = 488 $*{10}^{-9}$ m

Stopping potential of the photoelectrons, $V_0$ = 0.38 V = $\frac{0.38}{1.6*{10}^{-19}}$ $eV$

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

Charge of an electron, e = 1.6 $*{10}^{-19}$ C

Speed of the light, c = 3 $*{10}^8$ m/s

From Einstein's photoelectric effect, we have the relation involving the work function ${\varnothing }_0$ of the material of the emitter as:

$e{V}_0=\frac{hc}{\lambda }-{\varnothing }_0$

${\varnothing }_0=\frac{hc}{\lambda }-e{V}_0$ = $\frac{6.626\mathrm{}*{10}^{-34}*3\mathrm{}*{10}^8}{488\mathrm{}*{10}^{-9}*1.6*{10}^{-19}}$ $-$ 1.6 $*{10}^{-19}*\frac{0.38}{1.6*{10}^{-19}}$ = 2.54 – 0.38 = 2.165 eV

Therefore,, the material with which the emitter is made has the work function of 2.165 eV.

11.10 Frequency of the incident photon, $\nu =7.21*{10}^{14}$ Hz

Maximum speed of electron, v = 6 $*{10}^5$ m/s

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

Mass of electron, m = 9.1 $*{10}^{-31}$ kg

Let the threshold frequency = ${\nu }_0$

From the relation of threshold frequency and kinetic energy, we can write

$\frac{1}{2}$ m $v^2$ = $h$ ( $\nu -{\nu }_0)$

${\nu }_0=$ $\nu -\frac{m{v}^2}{2h}$ = $7.21*{10}^{14}-\frac{9.1\mathrm{}*{10}^{-31}*{(6\mathrm{}*{10}^5)}^2}{2*6.626\mathrm{}*{10}^{-34}}$ = 4.738 $*{10}^{14}$ Hz

Therefore, the threshold frequency is 4.738 $*{10}^{14}$ Hz.

11.9 Work function of the metal, ${\varnothing }_0$ = $4.2eV$

Charge of an electron, e = 1.6 $*{10}^{-19}$ C

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

Wavelength of the incident radiation, $\lambda$ = 330 nm= 330 $*{10}^{-9}$ m

Speed of the light, c = 3 $*{10}^8$ m/s

The energy of the incident photon is given as:

$E$ = $\frac{hc}{\lambda }$ = $\frac{6.626\mathrm{}*{10}^{-34}*3\mathrm{}*{10}^8}{330\mathrm{}*{10}^{-9}}$ = 6.02 $*{10}^{-19}$ J = $\frac{6.02\mathrm{}*{10}^{-19}}{1.6*{10}^{-19}}$ eV = 3.76 eV

Since the energy of the incident photon (3.76 $eV$ ) is less than the work function of the metal (4.2 $eV$ ), there will be no photoelectric emission.

11.8 Threshold frequency of the metal, ${\nu }_0=3.3*{10}^{14}$ Hz

Frequency of the light incident on metal, $\nu =8.2*{10}^{14}$ Hz

Charge of an electron, e = 1.6 $*{10}^{-19}$ C

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

Let the cut-off voltage for the photoelectric emission from the metal be $V_0$

The equation of the cut-off energy is given as:

$e{V}_0$ = $h(\nu -{\nu }_0)$ or

$V_0=\frac{h(\nu -{\nu }_0)}{e}$ = $\frac{6.626\mathrm{}*{10}^{-34}*(8.2*{10}^{14}-3.3*{10}^{14})}{1.6\mathrm{}*{10}^{-19}}$ V = 2.0292 V

Therefore, the cut-off voltage for the photoelectric emission is 2.0292V

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