Physics

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

Length of the mild steel wire, l = 1.0 m

Area of cross-section, A = 0.5 $*{10}^{-2}$ ${cm}^2$ = 0.5 $*{10}^{-6}{m}^2$

A mass of 100 gm is suspended at the midpoint.

m = 100 gm= 0.1 kg

Due to the weight, the wire dips, as shown in the figure.

Original length = XZ, depression = l

The final length of the wire after it dips = XO + OZ

Increase in length of the wire, Δl = (XO + OZ) – XZ ……(i)

From Pythagoras theorem

XO = OZ = $\sqrt{{0.5}^2+l^2}$

From equation (i)

Δl = 2 $*\sqrt{{0.5}^2+l^2}$ - 1.0 = 2 $*0.5*\sqrt{1+{\left(\frac{l}{0.5}\right)}^2}$ - 1.0 = $\sqrt{1+{\left(\frac{l}{0.5}\right)}^2}$ - 1.0

Neglecting the smaller terms, we can write, Δl = $\frac{l^2}{0.5}$

We know, Strain = $\frac{Increaseinlength}{Originallength}$

Let T be the tension in the wire, then

mg = 2T $\cos ?\theta$

From the figure

$\cos ?\theta$ = $\frac{OY}{OX}$ = $\frac{0.5}{\sqrt{{0.5}^2+l^2}}$ = $\frac{0.5}{0.5*\sqrt{1+{\left(\frac{l}{0.5}\right)}^2}}$

Expanding the expression and eliminating the higher terms:

$\cos ?\theta$ = $\frac{l}{0.5*\sqrt{1+{\left(\frac{1}{2}\right)\left(\frac{l}{0.5}\right)}^2}}$

Since ( 1+ $\frac{l^2}{0.5}$ ) = 1 for small l, $\cos ?\theta$ = $\frac{1}{0.5}$

Then T = $\frac{mg}{2\left(\frac{l}{0.5}\right)}$ = $\frac{mg*0.5}{2l}$ = $\frac{mg}{4l}$

Stress = $\frac{Tension}{Area}$ = $\frac{mg}{4l*A}$

Young's modulus = $\frac{Stress}{Strain}$

Y = $\frac{mg*0.5}{?l*A*l^2}$

l= $\sqrt{\frac{mg*0.5}{4YA}}$

For steel, Y = 2 $*{10}^{11}$ Pa, then

l = $\sqrt{\frac{0.1*9.8*0.5}{4*2*{10}^{11}*0.5*{10}^{-6}}}$ = 0.0106 m

Hence, the depression at the midpoint is 0.0106 m

Cross-sectional area of wire A, $a_1$ = 1 ${mm}^2$ = 1 $*{10}^{-6}{m}^2$

Cross-sectional area of wire B, $a_2$ = 2 ${mm}^2$ = 2 $*{10}^{-6}{m}^2$

Young's modulus for steel, $Y_1$ = 2 $*{10}^{11}$ N/ $m^2$

Young's modulus for aluminium, $Y_2$ = 7 $*{10}^{10}$ N/ $m^2$

Stress in the wire = $\frac{Force}{area}$ = $\frac{F}{a}$

If the two wires have equal stresses, then

$\frac{F_1}{a_1}$ = $\frac{F_2}{a_2}$ or $\frac{F_1}{F_2}$ = $\frac{a_1}{a_2}$ = $\frac{1}{2}$ ………(i)

Where $F_1$ is the force exerted on steel wire and $F_2$ is the force exerted on aluminium wire

Taking a moment around the point of suspension, we get

$F_{1}y$ = $F_2(1.05-y)$

$\frac{F_1}{F_2}$ = $\frac{(1.05-y)\mathrm{}}{y}$ ……(ii)

Using equation (i) and (ii), we can write

$\frac{1}{2}=\frac{(1.05-y)\mathrm{}}{y}$ or y = 2.1 – 2y or y = 2.1/3 = 0.7 m

We know Young's modulus Y = Stress / Strain

Y = (F/A)/ Strain or Strain = (F/A)/Y

In case of Steel wire, ${Strain}_{steel}$ = $\frac{F_1}{a_1*Y_1}$

In case of Aluminium wire, ${Strain}_{alu}$ = $\frac{F_2}{a_2*Y_2}$

Since both the strains are equal, we get

$\frac{F_1}{F_2}$ = $\frac{a_1*Y_1}{a_2*Y_2}$ = $\frac{1*{10}^{-6}*2*{10}^{11}}{2*{10}^{-6}*7*{10}^{10}}$ = 1.429

From (ii) we get

$\frac{(1.05-y)\mathrm{}}{y}$ = 1.429

y = 0.432m

In order to produce an equal strain in the two wires. The mass should be suspended at a distance of 0.432m from the end A