Class 12th

11.30 Intensity of the incident light, I = ${10}^{-5}$  W $m^{-2}$

Surface area of the sodium photocell, A = 2 ${cm}^2$  = 2 $*{10}^{-4}{m}^2$

The incident power of the light, P = I $*A$  = ${10}^{-5}$ $*{10}^{-4}{m}^2$ W = 2 $*{10}^{-9}$  W

Work function of the metal, ${\varnothing }_0$ = 2 eV = 2 $*1.6*{10}^{-19}$  J = 3.2 $*{10}^{-19}$  J

Number of layers of sodium that absorbs the incident energy, n = 5

The effective atomic area of a sodium atom, $A_e$  = ${10}^{-20}$ $m^2$

Hence, the number of conduction electrons in n layers is given by:

n' = n $*\frac{A}{A_e\mathrm{}}$  = 5 $*\frac{2*{10}^{-4}}{{10}^{-20}}$  = 1 $*{10}^{17}$

Since the incident power is uniformly absorbed by all the electrons continuously, the amount of energy absorbed per second per electron is

$E=\frac{P}{n\text{'}}$  = $\frac{2*{10}^{-9}}{1*{10}^{17}}$  = 2 $*{10}^{-26}$  J/s

Time required for photoelectric emission,

t = $\frac{{\varnothing }_0}{E}$  = $\frac{3.2*{10}^{-19}}{2*{10}^{-26}}$ s = 16 $*{10}^6$  seconds = 0.507 years

The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.

11.29 Wavelength of the radiation, $\lambda$ = 3300 Å = 3300 $*{10}^{-10}$ m

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

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

The energy of the incident radiation is given as:

$E$ = $\frac{hc}{\lambda }$

= $\frac{6.626*{10}^{-34}*3*{10}^8\mathrm{}}{3300*{10}^{-10}}$ = 6.024 $*{10}^{-19}$ J = $\frac{6.024*{10}^{-19}}{1.6*{10}^{-19}}$ eV = 3.765 eV

It can be observed that the energy of incident radiation is greater than the work function of Na and K only. Mo and Ni having more work function will not show the photoelectric emission.

If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase, but it will not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.

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.