Physics Ncert Solutions Class 12th

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

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.