Physics

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.

Let us assume the depth = h, pressure at depth, $P_h$ = 80 atm = 80 $*1.013*{10}^5$ Pa

Density of water at the surface, ${\rho }_s$ = 1.03 $*{10}^3$ kg/ $m^3$

Let density of water at depth h be ${\rho }_h$

Let $V_s$ be the volume at the surface and $V_h$ be the volume at depth h and ΔV be the change in volume. Let m be the mass of water.

ΔV = $V_s-V_h$ From the relation m = $\rho V,$ we get

ΔV = m ( $\frac{1}{{\rho }_s}$ - $\frac{1}{{\rho }_h}$ ) = ${\rho }_s{V}_s$ ( $\frac{1}{{\rho }_s}$ - $\frac{1}{{\rho }_h}$ )

$\frac{\mathrm{\Delta }\mathrm{V}}{V_s}$ = 1- $\frac{{\rho }_s}{{\rho }_h}$ ……(i)

Bulk modulus of water, k = $V_1\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{\Delta }\mathrm{V}}$ = $V_s\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{\Delta }\mathrm{V}}$

$\frac{\mathrm{\Delta }\mathrm{V}}{V_s}$ = $\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{k}}$ …….(ii)

Bulk modulus of water, k = 2.1 $*{10}^9$ Pa

Hence $\frac{\mathrm{\Delta }\mathrm{V}}{V_s}=\frac{80\mathrm{}*1.013*{10}^5}{2.1*{10}^9}$ = 3.86 $*{10}^{-3}$

From equation (i) $\frac{{\rho }_s}{{\rho }_h}$ = 0.99614

${\rho }_h=1.03*{10}^3/0.99614$ = 1.034 $*{10}^3$ kg/ $m^3$