Ncert Solutions Physics Class 12th

11.28 Einstein's photoelectric equation is given as:

$e{V}_0$ = $h\nu -{\varnothing }_0$

$V_0=\frac{h}{e}\nu -\frac{{\varnothing }_0}{e}$ ……….(1)

where

$V_0=$ Stopping potential

h = Planck's constant

e = Charge of an electron

$\nu =\mathrm{F}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$

${\varnothing }_0=\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}$

Speed of light, c = 3 $*{10}^8$ m/s

It can be concluded from equation (1) that $V_0$ is directly proportional to frequency $\nu$

Now frequency $\nu$ can be expressed as

$\nu =\frac{c}{\lambda }$

Then,

  ${\nu }_1$ = $\frac{c}{{\lambda }_1}$ = $\frac{3*{10}^8}{3650\mathrm{}\mathrm{\AA }}$ Hz = 8.219$*{10}^{14}$ Hz

 ${\nu }_2$ = $\frac{c}{{\lambda }_2}$ = $\frac{3*{10}^8}{4047\mathrm{}\mathrm{\AA }}$  =  $\frac{3*{10}^8}{3650*{10}^{-10}\mathrm{}}$Hz = 7.413$*{10}^{14}$
 Hz

${\nu }_3$ = $\frac{c}{{\lambda }_3}$ = $\frac{3*{10}^8}{4358\mathrm{}\mathrm{\AA }}$ =$\frac{3*{10}^8}{4047*{10}^{-10}}$Hz = 6.883$*{10}^{14}$  Hz

${\nu }_4$ = $\frac{c}{{\lambda }_4}$ = $\frac{3*{10}^8}{5461\mathrm{}\mathrm{\AA }}$ =$\frac{3*{10}^8}{4358*{10}^{-10}}$Hz = 5.493$*{10}^{14}$ $\frac{\mathrm{}}{}$ Hz

${\nu }_5$ = $\frac{c}{{\lambda }_5}$ = $\frac{3*{10}^8}{6907\mathrm{}\mathrm{\AA }}$ =$\frac{3*{10}^8}{5461*{10}^{-10}}$Hz = 4.343$*{10}^{14}$Hz

From the given data of stopping potential, we get the following table

It can be observed that the obtained curve is a straight line. It intersects the frequency axis at 5 $*{10}^{14}$ Hz, which is the threshold frequency (${\nu }_0$
 ) of the material. Point D corresponds to a frequency less than threshold frequency. Hence, there is no photoelectric emission for the${\lambda }_5$ line and therefore, no stopping voltage is required to stop the current.

Slope of the straight line = $\frac{AB}{BC}$ = $\frac{1.28-0.16}{(8.219-5.493)*{10}^{14}}$

From equation (1), the slope $\frac{h}{e}$
 can be written as

 $\frac{h}{e}=\frac{1.28-0.16}{(8.219-5.493)*{10}^{14}}$ = $\frac{1.12}{2.726*{10}^{14}}$ Js

 $h$ = $\frac{1.12}{2.726*{10}^{14}}*1.6*{10}^{-19}$ = 6.573 $*{10}^{-34}$ Js

The work function of the material is given by, ${\varnothing }_0$  = $h{\nu }_0$  = 6.573 $*{10}^{-34}*$ 5 $*{10}^{14}$

= 3.286 $*{10}^{-19}$  J = $\frac{3.286*{10}^{-19}}{1.6*{10}^{-19}}$ eV = 2.054eV

11.27 Wavelength of the monochromatic light, $\lambda$ = 640.2 nm = 640.2 ${*10}^{-9}m$

Stopping potential of neon lamp, $V_0$ = 0.54 V

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

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

Speed of light, c = 3 $*{10}^8$ m/s

Let ${\varnothing }_0$ be the work function and $\nu$ frequency of emitted light

We have the photo-energy relation from the photoelectric effect as:

$e{V}_0$ = $h\nu -{\varnothing }_0$ = h $\frac{c}{\lambda }$ - ${\varnothing }_0$

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

= $\frac{6.626*{10}^{-34}*3*{10}^8}{640.2{*10}^{-9}}$ $-$ 1.6 $*{10}^{-19}*0.54$

= 3.105 $*{10}^{-19}$ - 0.864 $*{10}^{-19}$

= 2.241 $*{10}^{-19}$ J

= $\frac{2.241*{10}^{-19}}{1.6*{10}^{-19}}$ eV

=1.40 eV

The wavelength of the radiation emitted from an iron source, $\lambda \text{'}$ = 427.2 nm = 427.2 $*{10}^{-9}$ m

Let the new stopping potential be $V_0\text{'}$

From the relation ${\varnothing }_0=\frac{hc}{\lambda \text{'}}$ - $e{V}_0\text{'}$ , we get

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

= $\frac{6.626*{10}^{-34}*3*{10}^8}{427.2*{10}^{-9}}$ - 2.241 $*{10}^{-19}$

= 4.653 $*{10}^{-19}$ - 2.241 $*{10}^{-19}$

= 2.412 $*{10}^{-19}$ J

= $\frac{2.412*{10}^{-19}}{1.6*{10}^{-19}}$ eV

= 1.51 eV

The new stopping potential is 1.51 eV.

11.26 Wavelength of ultraviolet light, $\lambda$ = 2271Å = 2271 $*{10}^{-10}$ m

Stopping potential of the metal, $V_0$ = 1.3 V

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

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

Speed of light, c = 3 $*{10}^8$ m/s

Work function of the metal, ${\varnothing }_0$

Frequency of light = $\nu$

We have the photo-energy relation from the photoelectric effect as:

${\varnothing }_0=h\nu -e{V}_0$

= $\frac{hc}{\lambda }-e{V}_0$

= $\frac{6.626*{10}^{-34}*3*{10}^8}{2271*{10}^{-10}}-1.6*{10}^{-19}*1.3$

= 8.75 $*{10}^{-19}$ - 2.08 $*{10}^{-19}$

= 6.67 $*{10}^{-19}$ J

= $\frac{6.67*{10}^{-19}}{1.6*{10}^{-19}}$ eV

= 4.17 eV

Let ${\nu }_0$ be the threshold frequency of the metal.

Therefore, ${\varnothing }_0$ = h ${\nu }_0$

${\nu }_0=$ $\frac{{\varnothing }_0}{h}$ = $\frac{6.67*{10}^{-19}}{6.626*{10}^{-34}}$ Hz = 1.007 $*{10}^{15}$ Hz

Wavelength of red light, ${\lambda }_r$ = 6328 Å = 6328 $*{10}^{-10}$ m

Frequency of red light, ${\nu }_r$ = $\frac{c}{{\lambda }_r}$ = $\frac{3*{10}^8}{6328*{10}^{-10}}$ = 4.74 $*{10}^{14}$ Hz

Since ${\nu }_0>{\nu }_r,\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}\mathrm{}\mathrm{w}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{t}\mathrm{o}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{l}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{b}\mathrm{y}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{r}.$

11.25 The power of the medium wave transmitter, P = 10 kW = 10 $*{10}^3$ W = ${10}^4$ J/s

Hence energy emitted by the transmitter per second, E = ${10}^4$ J

Wavelength of the radio wave, $\lambda$ = 500 m

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

Speed of light, c = 3 $*{10}^8$ m/s

Energy of the wave is given as :

$E_w$ = $\frac{hc}{\lambda }$ = $\frac{6.626*{10}^{-34}*3*{10}^8}{500}$ = 3.98 $*{10}^{-28}$ J

Let n be number of photons emitted by the transmitter. Hence, total energy transmitted is given by:

n $E_w$ = E

n = $\frac{E}{E_w}$ = $\frac{{10}^4}{3.98*{10}^{-28}}$ = 2.52 $*{10}^{31}$

Intensity of light perceived by the human eye, I = ${10}^{-10}$ W $m^{-2}$

Area of the pupil, A = 0.4 ${cm}^2$ = 0.4 $*{10}^{-4}{m}^2$

Frequency of white light, $\nu$ = 6 $*{10}^{14}$ Hz

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

Energy emitted by photon is given as:

$E=h\nu$ = 6.626 $*{10}^{-34}*6*{10}^{14}$ = 3.9756 $*{10}^{-19}$ J

If n is total number of photons falling per second, per unit area of the pupil,

Total energy per unit area of n photons is = n $*$ 3.9756 $*{10}^{-19}$ J = Intensity of light

n $*$ 3.9756 $*{10}^{-19}$ = ${10}^{-10}$

n = 2.52 $*{10}^8$

Total number of photons entering the pupil per second is given as:

$n_A$ = n $*A$ = 2.52 $*{10}^8*$ 0.4 $*{10}^{-4}$ = 10.06 $*{10}^3$ /s

The number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.