5 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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Question 15 Marks

In an experiment on photoelectric effect, the stopping potential is measured for monochromatic light beams corresponding to different wavelengths. The data collected are $11s$ follows:

wavelength $(nm)$	$350$	$400$	$450$	$500$	$550$
stopping potential$(V):$	$1.45$	$1.00$	$0.66$	$0.38$	$0.16$

Plot the stopping potential against inverse of wavelength $\big(\frac{1}{\lambda}\big) $ on a graph paper and find

The Planck constant,
The work function of the emitter and.
The threshold wavelength.

Answer

when $\lambda=350,\text{v}_\text{s}=1.45$

and when $\lambda=400,\text{v}_\text{s}=1$
$\therefore\frac{\text{hc}}{350}=\text{w+1.45}\dots (1)$
and $\frac{\text{hc}}{400}=\text{w+1}\dots(2)$
Subtracting $(2)$ from $(1)$ and solving to get the value of $h$ we get
$h = 4.2 \times 10^{-15} ev-\sec$

Now work function $=\text{w}=\frac{\text{hc}}{\lambda}=\text{ev-s}$

$=\frac{1240}{350}-1.45=2.15\ \text{ev.}$

$\text{w}=\frac{\text{hc}}{\lambda}=\lambda_{\text{there cathod}}=\frac{\text{hc}}{\text{w}}$

$=\frac{124}{2.15}=576.8\ \text{nm.}$

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Question 25 Marks

It is found that yellow light does not eject photoelectrons from a metal. Is it advisable to try with orange light? With green light?

Answer

Photoelectrons are emitted from a metal's surface if the frequency of incident radiation is more than the threshold frequency of the given metal surface. As yellow light does not eject photoelectrons from a metal it means that the threshold frequency of the metal is more than the frequency of yellow light. Since the frequency of orange light is less than the frequency of yellow light, therefore it will not be able to eject photoelectrons from the metal's surface. The frequency of green light is more than the frequency of yellow light. Hence, when it is incident on the metal surface, it will eject electrons from the metal.

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Question 35 Marks

A horizontal cesium plate ($\phi$ = l.9 eV) is moved vertically downward at a constant speed u in a room full of radiation of wavelength 250 run and above. What should be the minimum value of u so that the vertically upward component of velocity is nonpositive for each photoelectron?

Answer

Work function $=\phi,$ distance = dThe particle will move in a circle
When the stopping potential is equal to the potential due to the singly charged ion at that point.
$\text{eV}_0=\frac{\text{hc}}{\lambda}-\phi$
$\Rightarrow\text{v}_0=\Big(\frac{\text{hv}}{\lambda}-\phi\Big)\frac{1}{\text{e}}$
$\Rightarrow\frac{\text{ke}}{2\text{d}}=\Big(\frac{\text{hc}}{\lambda}-\phi\Big)\frac{1}{\text{e}}$
$\Rightarrow\frac{\text{ke}}{\text{2d}}=\frac{\text{hc}}{\lambda}-\phi$
$\Rightarrow\frac{\text{hc}}{\lambda}=\frac{\text{ke}^2}{2\text{d}}+\phi=\frac{\text{ke}^2+2\text{d}\phi}{2\text{d}}$
$\Rightarrow\lambda=\frac{\text{hc 2d}}{\text{ke}^2+2\text{d}\phi}=\frac{2\text{hcd}}{\frac{1}{4\pi\omega_0\text{e}^2}2\text{d}\phi}=\frac{8\pi\omega_0\text{hcd}}{\text{e}^2+8\pi\omega_0\text{d}\phi}$

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Question 45 Marks

A small piece of cesium metal $(\phi = 1 ·9 eV)$ is kept at a distance of $20\ cm$ from a large metal plate having a charge density of $1.0 \times 10^{-9}C/m^2$ on the surface facing the cesium piece. A monochromatic light of wavelength $400\ nm$ is incident on the cesium piece. Find the minimum and the maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.

Answer

Given $\sigma = 1 \times 10^{-9}cm^{-2},$
$W_0 (Cs) = 1.9eV, $
$d = 20\ cm = 0.20m,$
$\lambda = 400\ nm$
we know $\rightarrow$ Electric potential due to a charged plate $= V = E \times d$
Where $E \rightarrow$ elelctric field due to the charged plate $=\frac{\sigma}{\text{E}_0}$
$d \rightarrow$ Separation between the plates.
$\text{v}=\frac{\sigma}{\text{E}_0}\times\text{d}=\frac{1\times10^{-9}\times20}{8.85\times10^{-12}\times100}$
$=22.598\text{V}=22.6$
$\text{V}_0\text{e}=\text{hv}-\text{w}_0=\frac{\text{hc}}{\lambda}-\text{w}_0$
$=\frac{4.14\times10^{-15}\times3\times10^{8}}{4\times10^{-7}}-1.9$
$=3.105-1.9=1.205\text{ev.}$
$V_0 = 1.205V$
As $V_0$ is much less than $'V\ ’$
Hence the minimum energy required to reach the charged plate must be $= 22.6eV$
For maximum $KE,$ the $V$ must be an accelerating one.
Hence max $KE = V_0 + V = 1.205 + 22.6 = 23.8005ev$

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Question 55 Marks

Consider the situation of the previous problem. Consider the fastest electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate.

Answer

Here electric field of metal plate $=\text{E}=\frac{\text{p}}{\text{E}_0}$
$=\frac{1\times10^{-19}}{8.85\times10^{-12}}=113\text{v/m}$
accl. de $=\phi=\frac{\text{qE}}{\text{m}}$
$=\frac{1.6\times10^{-19}\times113}{9.1\times10^{-31}}=19.87\times10^{12}$
$\text{t}=\frac{\sqrt{2\text{y}}}{\text{a}}=\frac{\sqrt{2\times20\times10^{-2}}}{19.87\times10^{-31}}=1.41\times10^{-7}\text{sec}$
$K.E. =\frac{\text{hc}}{\lambda}-\text{w}=1.2\text{ev} = 1.2 \times 1.6 \times 10 - 19J [$because in previous problem i.e. in problem $31 : KE = 1.2ev]$
$\therefore\text{v}=\frac{\sqrt{2\text{KE}}}{\text{m}}$
$=\frac{\sqrt{2\times1.2\times1.6\times10^{-19}}}{4.1\times10^{-31}}=0.665\times10^{-6}$
$\therefore$ Horizontal displacement $= Vt \times t$
$= 0.655 \times 10^{-6} \times 1.4 \times 10^{-7} = 0.092m = 9.2\ cm.$

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Question 65 Marks

What is the speed of a photon with respect to another photon if:

The two photons are going in the same direction and.
They are going in opposite directions?

Answer

In relativity, the relative speed of two objects $(\text{v}_\text{rel})$ moving in the same direction with speeds u and v is given by.

$\text{v}_\text{rel}=\frac{\text{u-v}}{1-\frac{\text{uv}}{\text{c}^2}}\dots(1)$
As the photons are moving with the speed of light, u = c and v = c.
On substituting the values of u and v in equation (1), we get:
$\text{v}_\text{rel}=0$
Thus, relative velocity of a photon with respect to another photon will be 0, when they are going in the same direction.

In relativity, relative speed of two objects moving in opposite directions with speeds u and v is given by.

$\text{v}_\text{rel}=\frac{\text{u+v}}{1+\frac{\text{uv}}{\text{c}^2}}\dots(2)$
We know that a photon travels with the speed of light. Therefore, u = c and v = c
On substituting the values of u and v in equation (2), we get:
$\text{v}_\text{rel}=\text{c}$
Thus, the relative velocity of a photon with respect to another photon will be equal to the speed of light when they are going in opposite directions.

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Question 75 Marks

A light beam of wavelength $400$ run is incident on a metal plate of work function $2.2eV.$

A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that $10\%$ of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal.
Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.

Answer

When $\lambda=400\text{nm}$

Energy of photon
$=\frac{\text{hc}}{\lambda}=\frac{1240}{400}=3.1\text{ev.}$
This energy given to electron
But for the first collision energy lost $= 3.1ev \times 10\% = 0.31ev$
for second collision energy lost $= 3.1ev \times 10\% = 0.31 ev$
Total energy lost the two collision $= 0.31 + 0.31 = 0.62ev$
$K.E.$ of photon electron when it comes out of metal
$=\frac{\text{hc}}{\lambda} -$ work function $-$ Energy lost due to collision
$= 3.1ev - 2.2 - 0.62 = 0.31ev$

For the $3^{rd}$ collision the energy lost $= 0.31ev$

Which just equative the $KE$ lost in the $3^{rd}$ collision electron. It just comes out of the metal
Hence in the fourth collision electron becomes unable to come out of the metal
Hence maximum number of collision $= 4.$

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Question 85 Marks

In the arrangement shown in figure $y = 1.0\ mm. d = 0.24\ mm$ end $D = 1.2m.$ The work function of the materiel of the emitter is $2.2eV.$ Find the stopping potential V needed to stop the photocurrent.

Answer

Given : fringe width, $y = 1.0\ mm \times 2 = 2.0\ mm,$
$D = 0.24\ mm, W_0 = 2.2ev,$
$D = 1.2m\text{y}=\frac{\lambda\text{D}}{\text{d}}$
$\lambda=\frac{\text{yd}}{\text{D}}=\frac{2\times10^{-3}\times0.24\times10^{-3}}{1.2}=4\times10^{-7}\text{m}$
$\text{E}=\frac{\text{hc}}{\lambda}=\frac{4.14\times10^{-15}\times3\times10^{8}}{4\times10}=3.105\text{ev.}$
Stopping potential $eV_0 = 3.105 - 2.2 = 0.905V$

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Question 95 Marks

Figure is the plot of the stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find.

The ratio h/e and.
The work function.

Answer

We have to take two cases:
Case $I: v_0 = 1.656 v = 5 \times 10^{14} Hz$
Case $II: v_0 = 0 v = 1 \times 10^{14}Hz$ We know;

$ev_0 = hv - w_0$

$1.656e = h \times 5 \times 1014 - w_0 …(1)$
$0 = 5h \times 10^{14 -} 5w_0 …(2)$
$1.656e = 4w_0$
$\Rightarrow\text{w}_0=\frac{1.6556}{4}\text{ev}=0.414\text{ev}$

Putting value of w0 in equation $(2)$

$\Rightarrow 5w_0 = 5h \times 10^{14}$
$\Rightarrow 5 \times 0.414 = 5 \times h \times 10^{14}$
$\Rightarrow h = 4.414 \times 10 - 15 ev-s$

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Question 105 Marks

Consider the de$-$Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have.

The same speed.
The same momentum.
The same energy?.

Answer

de$-$Broglie wavelength$,\lambda=\frac{\text{h}}{\text{mv}},$
where $h =$ Planck's constant
$m =$ mass of the particle
$v =$ velocity of the particle

It is given that the speed of an electron and proton are equal.

It is clear from the above equation that
$ \lambda\alpha\frac{1}{\text{m}}$
As mass of a proton$, m_p >$ mass of an electron$, m_e,$ the proton will have a smaller wavelength compared to the electron.

$\lambda=\frac{\text{h}}{\text{p}}(\therefore\text{p}=\text{mv})$

So, when the proton and the electron have same momentum, they will have the same wavelength.

De$-$Broglie wavelength $(\lambda)$ is also given by

$\lambda=\frac{\text{h}}{\sqrt{2\text{mE}}},$
where $E =$ energy of the particle.
Let the energy of the proton and electron be $E.$
Wavelength of the proton,
$\lambda=\frac{\text{h}}{\sqrt{2\text{m}_\text{p}\text{E}}}\dots(1)$
Wavelength of the electron,
$\lambda=\frac{\text{h}}{\sqrt{2\text{m} _\text{e}\text{E}}}\dots (2)$
Dividing $(2)$ by $(1),$ we get:
$\frac{\lambda_\text{e}}{\lambda_\text{p}}=\frac{\sqrt{\text{m}_\text{e}}}{\sqrt{\text{m}}_\text{p}}$
$\Rightarrow\frac{\lambda_e}{\lambda_\text{p}}<1$
$\Rightarrow\lambda_\text{e}<\lambda_\text{p}$
It is clear that the proton will have smaller wavelength compared to the electron.

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Question 115 Marks

A horizontal cesium plate ($\phi$ = 1.9eV) is moved vertically downward at a constant speed u in a room full of radiation of wavelength 250 run and above. What should be the minimum value of u so that the vertically upward component of velocity is nonpositive for each photoelectron?

Answer

When $\lambda=250\text{nm}$Energy of photon
$=\frac{\text{hc}}{\lambda}=\frac{1240}{250}=4.96\text{ev}$
$\therefore\text{K.E.}=\frac{\text{hc}}{\lambda}-\text{w}=4.96-1.9\text{ev.}$
Velocity to be non positive for each photo electron
The minimum value of velocity of plate should be = velocity of photo electron
$\therefore$ Velocity of photo electron $=\sqrt{\frac{2\text{KE}}{\text{m}}}$
$=\sqrt{\frac{3.06}{9.1\times10^{-31}}}$
$=\sqrt{\frac{3.06\times1.6\times10^{-19}}{9.1\times10^{-31}}}=1.04\times10^6\text{m/sec.}$

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Question 125 Marks

The electric field associated with a light wave is given by $\text{E}=\text{E}_0\sin[(1.57\times10^7\text{m}^{-1})(\text{x}-{\text{ct}})].$ Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function $1.9\ \ce{ev}.$

Answer

$\text{E}=\text{E}_0\sin[(1.57\times10^{7}\text{m}^{-1})(\text{x}-\text{ct})]$
$\text{w}=1.57\times10^7\times\text{c}$
$\Rightarrow\text{f}=\frac{1.57\times10^{7}\times3\times10^{8}}{\text{2}\pi}\text{Hz}$
$\text{w}_0=1.9\ \text{ev}$
Now $eV_0 = hv - W_0$
$=4.14\times10^{-15}\times\frac{1.57\times3\times10^{15}}{2\pi}-1.9\ \text{ev}$
$=3.105-1.9=1.205\ \text{ev}$
So, $\text{v}_0=\frac{1.205\times1.6\times10^{-19}}{1.6\times10^{-19}}$
$=1.205\ \text{ev.}$

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Question 135 Marks

In en experiment on photoelectric effect, light of wavelength $400$ run is incident on a cesium plate at the rate of $5.0W.$ The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current reaches its saturation value. Assuming that on the average one out of every $10^\circ$ photons is able to eject a photoelectron, find the photocurrent in the circuit.

Answer

$\lambda=400\ \text{nm},\text{p}=5\text{w}$
$E$ of $1$ photon $=\frac{\text{hc}}{\lambda}=\Big(\frac{1242}{400}\Big)\text{ev}$
No.of electrons
$=\frac{5}{\text{Energy of 1 photon}}$
$=\frac{5\times400}{1.6\times10^{-19}\times1242}$
No.of electrons $= 1$ per $10^6$ photon.
No.of photoelectrons emitted
$=\frac{5\times400}{1.6\times1242\times10^{-19}\times10^{6}}$
Photo electric current
$=\frac{5\times400}{1.6\times1242\times10^{6}\times10^{-19}}\times1.6\times10^{-19}$
$=1.6\times10^{-6}\text{A}$
$=1.6\ \mu\text{A}.$

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Question 145 Marks

A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10W of power, find the force exerted by it on the surface.

Answer

Power = 10W P → Momentum$\lambda=\frac{\text{h}}{\text{p}}$
$\text{p}=\frac{\text{h}}{\lambda}$
$\frac{\text{p}}{\text{t}}=\frac{\text{h}}{\lambda\text{t}}$
$\text{E}=\frac{\text{hc}}{\lambda}$
$\frac{\text{E}}{\text{t}}=\frac{\text{hc}}{\lambda\text{t}}=\text{power(W)}$
$\text{W}=\frac{\text{pc}}{\text{t}}$
$\frac{\text{p}}{\text{t}}=\frac{\text{W}}{\text{c}}=\text{force.}$
$\text{force}=\frac{7}{10}(\text{absorbed})+2\times\frac{3}{10}\text{(reflected)}$
$=\frac{7}{10}\times\frac{\text{W}}{\text{C}}+2\times\frac{3}{10}\times\frac{\text{W}}{\text{C}}$
$\Rightarrow\frac{7}{10}\times\frac{10}{3\times10^8}+2\times\frac{3}{10}\times\frac{10}{3\times10^8}$
$=\frac{13}{3}\times10^{-8}=4.33\times10^{-8}\text{N.}$

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Question 155 Marks

The work function of a photoelectric material is 4.0eV.

What is the threshold wavelength?
Find the wavelength of light for which the stopping potential is 2.5V.

Answer

$\phi=4\text{ev}=4\times1.6\times10^{-19}\text{J}$

Threshold wavelength $=\lambda$

$\phi=\frac{\text{hc}}{\lambda}$
$\Rightarrow\lambda=\frac{\text{hc}}{\phi}=\frac{6.63\times10^{-13}\times3\times10^8}{4\times1.6\times10^{-19}}$
$=\frac{6.63\times2}{6.4}\times\frac{10^{-27}}{10^{-9}}=3.1\times10^{-7}\text{m}=310\text{nm}.$

Stopping potential is 2.5V

$\text{E}=\phi+\text{eV}$
$\Rightarrow\frac{\text{hc}}{\lambda}=4\times1.6\times10^{-19}+1.6\times10^{-19}\times2.5$
$\Rightarrow\lambda=\frac{6.63\times10^{-34}\times3\times10^8}{\lambda\times1.6\times10^{-19}}=4+2.5$
$\Rightarrow\frac{6.63\times10^{-34}\times3\times10^8}{\lambda\times1.6\times10^{-19}\times6.5}=1.9125\times10^{-7}=190\text{nm.}$

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Question 165 Marks

In an experiment on photoelectric effect, the emitter and the collector plates are placed et a separation of 10cm and are connected through en ammeter without any cell A magnetic field B exists parallel to the plates. The work function of the emitter is 2.39eV and the light incident on it has wavelengths between 400nm and 600nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge.

Answer

$\phi_0=2.39\text{ev}$$\lambda_1=400\text{nm},\lambda_2=600\text{nm}$
for B to the minimum energy should be maximum
$\therefore\lambda$ should be minimum.
$\text{E}=\frac{\text{hc}}{\lambda}-\phi_0=3.105-2.39=0.715\text{ev.}$
The presence of magnetic field will bend the beam there will be no current if the electron does not reach the other plates.
$\text{r}=\frac{\text{mv}}{\text{qB}}$
$\Rightarrow\text{r}=\frac{\sqrt{2\text{mE}}}{\text{qB}}$
$\Rightarrow0.1=\frac{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times0.715}}{1.6\times10^{-19}\times\text{B}}$
$\Rightarrow\text{B}=2.85\times10^{-5}\text{T}$

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Question 175 Marks

The electric field at a point associated with a light wave is $\text{E}=\big(100\frac{\text{v}}{\text{m}}\big)\sin[(3.0\times10^{15}\text{s}^{-1})\text{t]sin[(6.0}\times10^{15}\text{s}^{-1})\text{t}].$ If this light falls on a metal surface having a work function of $2.0eV,$ what will be the maximum kinetic energy of the photoelectrons?

Answer

$\text{E}=100\sin[(3\times10^{15}\text{s}^{-1})\text{t}]-\cos[3\times10^{15}\text{s}^{-1} )\text{t]}$
$=100\frac{1}{2}[\cos[9\times10^{15}\text{s}^{-1})\text{t}]-\cos[3\times10^{15}\text{s}^{-1})\text{t}]$
The w are $9 \times 10^{15}$ and $3 \times 10^{15 }$ for largest $K.E.$
$\text{f}_\text{max}=\frac{\text{W}_{\text{max}}}{2\pi}=\frac{9\times10^{15}}{2\pi}$
$\text{E}-\phi=\text{K.E}.$
$\Rightarrow\text{hf}-\phi_0=\text{K.E}.$
$\Rightarrow\frac{6.63\times10^{-34}\times9\times10^{15}}{2\pi\times1.6\times10^{-19}}-2=\text{KE}$
$\Rightarrow\text{KE}=3.938\text{ev}=3.93\text{ev}.$

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Question 185 Marks

When the sun is directly overhead, the surface of the earth receives $1.4 \times 10^3 W m^{-2}$ of sunlight. Assume that the light is monochromatic with average wavelength $500nm$ and that no light is absorbed in between the sun and the earth's surface. The distance between the sun and the earth is $1.5 \times 10^{11}m.$

Calculate the number of photons falling per second on each square metre of earth's surface directly below the sun.
How many photons are there in each cubic metre near the earth's surface at any instant?
How many photons does the sun emit per second?

Answer

Here intensity $=\text{I}=1.4\times10^3\omega/\text{m}^2$

Let no.of photons/sec emitted $= n$
No.of photons$/m^2 =\frac{\text{nhc}}{\lambda} =$ intensity
Intensity,$\text{I}=\frac{\text{power}}{\text{area}}=1.4\times10^3\omega/\text{m}^2$
$\therefore$ Power $=$ Energy emitted/sec
$=\frac{\text{nhc}}{\lambda}=\text{p}$
$\text{n}=\frac{\text{intensity}\times\lambda}{\text{hc}}$
$=\frac{1.9\times10^3\times5\times10^{-9}}{6.63\times10^{-34}\times3\times10^8}=3.5\times10^{21}$

Consider no.of two parts at a distance $r$ and $r + dr$ from the source.

The time interval $‘dt’$ in which the photon travel from one point to another $=\frac{\text{dv}}{\text{e}}=\text{dt.}$
In this time the total no.of photons emitted $=\text{N}=\text{n}\text{ dt}=\Big(\frac{\text{p}\lambda}{\text{hc}}\Big)\frac{\text{dr}}{\text{C}}$
These points will be present between two spherical shells of radii $‘r’$ and $r+dr.$ It is the distance of the $1^{st}$ point from the sources. No.of photons per volume in the shell
$(\text{r+r+dr})=\frac{\text{N}}{2\pi\text{r2dr}}$
$=\frac{\text{p}\lambda\text{dr}}{\text{hc}^2}=\frac{1}{4\pi\text{r}^2\text{ch}}=\frac{\text{p}\lambda}{4\pi\text{hc}^2\text{r}^2}$
In the case $=1.5\times10^{11}\text{m},\lambda=500\text{nm},=500\times10^{-9}\text{m}$
$\frac{\text{p}}{4\pi^2}=1.4\times10^3,$
$\therefore$ No.of photons$/m^{3 }=\frac{\text{p}}{4\pi\text{r}^2}\frac{\lambda}{\text{hc}^2}$
$=1.4\times10^3\times\frac{500\times10^{-9}}{6.63\times10^{-34}\times10^8}=1.2\times10^{13}$

No.of photons $= ($No.of photons$/\sec/m^2) \times$ Area

$=(3.5\times10^{21})\times4\pi^2$
$=3.5\times10^{21}\times4(3.14) (1.5\times10^{11})^{12}.$
$=9.9\times10^{44}$

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Question 195 Marks

When a metal plate is exposed to a monochromatic beam of light of wavelength 400nm, a negative potential of 1.1V is needed to stop the photocurrent. Find the threshold wavelength for the metal.

Answer

$\lambda=400\text{nm}=400\times10^{-9}\text{m}$$\text{v}_0=1.1\text{v}$
$\frac{\text{hc}}{\lambda}=\frac{\text{hc}}{\lambda_0}+\text{ev}_0$
$\Rightarrow\frac{6.63\times10^{-34}\times3\times10^8}{400\times10^{-9}}$
$=\frac{6.63\times10^{-34}\times3\times10^{8}}{\lambda_0}+1.6\times10^{-19}\times1.1$
$\Rightarrow4.97=\frac{19.89\times10^{-26}}{\lambda_0}+1.76$
$\Rightarrow\frac{19.89\times10^{-26}}{\lambda_0}=4.97-17.6=3.21$
$\lambda_0=\frac{19.89\times10^{-28}}{3.21}=6.196\times10^{-7}\text{m}=620\text{nm.}$

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Question 205 Marks

Find the maximum magnitude of the linear momentum of a photoelectron emitted when light of wavelength 400 run falls on a metal having work function 2.5eV.

Answer

Energy of photoelectron$\Rightarrow\frac{1}{2}\text{mv}^2=\frac{\text{hc}}{\lambda}-\text{hv}_0$
$=\frac{4.14\times10^{-15}\times3\times10^{8}}{4\times10^{-7}}-2.5\text{ev}=0.605\text{ev.}$
We know KE $=\frac{\text{p}^2}{2\text{m}}\Rightarrow\text{p}^2=2\text{m}\times\text{KE.}$
$\text{p}^2=2\times9.1\times10^{-31}\times0.605\times1.6\times10^{-19}$
$\text{p}=4.197\times10^{-25}\text{kg}-\text{m/s}$

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