1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point 1. Xtra Edge February 2012 - Career Point
Failure of wave theory : Wave theory of light could not explain the laws of photoelectric effect. According to wave theory, the kinetic energy of the emitted electrons should increase with the increase of intensity of incident radiation. Kinetic energy of the emitted electron does not depend on the frequency of incident radiation according to wave theory. Wave theory failed to explain the existence of threshold frequency. According to wave theory there must be a time lag between the arrival of light and emission of photoelectrons. Einstein's theory of photoelectric effect : Einstein explained the laws of photoelectric effect on the basis of Planck's quantum theory of radiation. Einstein treated photoelectric effect as a collision between a photon and an atom in which photon is absorbed by the atom and an electron is emitted. According to law of conservation of energy, 1 2 hν = hν0 + mv 2 where hν is the energy of the incident photon; hv0 is the minimum energy required to detach the electron from the atom (work function or ionisation energy) and (1/2) mv 2 is the kinetic energy of the emitted electron. The above equation is known as Einstein's photoelectric equation. Kinetic energy of the emitted electron, 1 2 = mv = h(ν – ν0) = hν – W 2 Explanation of laws of photoelectric effect : (a) The KE of the emitted electron increases with the increase of frequency of incident radiation since W (work function) is constant for a given emitter. KE is directly proportional to (ν – ν0) (b) Keeping the frequency of incident radiation constant if the intensity of incident light is increased, more photons collide with more atoms and more photoelectrons are emitted. The KE of the emitted electron remains constant since the same photon collides with the same atom (i.e., the nature of the collision does not change). With the increase in the intensity of incident light photoelectric current increases. (c) According to Einstein's equation, if the frequency of incident radiation is less than certain minimum value, the photoelectric emission is not possible. This frequency is known as threshold frequency. Hence, the frequency of incident radiation below which photoelectric emission is not possible is known as threshold frequency or cut-off frequency. It is given by : hν − ( 1/ 2) mv ν0 = h On the other hand, if the wavelength of the incident radiation is more than certain critical value, then photoelectric emission is not possible. This wavelength is known as threshold wavelength of cut-off wavelength. It is given by : hc λ0 = 2 [ hν − ( 1/ 2) mv ] (d) Since Einstein treated photoelectric effect as a collision between a photon and an atom, he explained the instantaneous nature of photoelectric effect. Some other important points : Stopping potential : The negative potential applied to the collector in order to prevent the electron from reaching the collector (i.e., to reduce the photoelectric current to zero) is known as stopping potential. 1 2 eV0 = mvmax. = hν – W = h(ν – ν0) 2 Millikan measured K.E. of emitted electrons or stopping potentials for different frequencies of incident radiation for a given emitter. He plotted a graph with the frequency on x-axis and stopping potential on y-axis. The graph so obtained was a straight line as shown in figure. ν0 Frequency of incident light XtraEdge for IIT-JEE 22 FEBRUARY 2012 V0(stopping potential) 2 Millikan measured the slope of the straight line (=h/e) and calculated the value of Planck's constant. I Full intensity 75% intensity 50% intensity 25% intensity V0 – + Potential difference
The intercept of V0 versus ν graph on frequency axis is equal to threshold frequency (ν0). From this, the work function (hν0) can be calculated. Graphs in photoelectric effect : (a) Photoelectric current versus potential difference graphs for varying intensity (keeping same metal plate and same frequency of incident light) : These graphs indicate that stopping potential is independent of the intensity and saturation current is directly proportional to the intensity of light. ν2>ν1 ν2 ν1 I – (V0) 2 (V0) 1 Potential difference + (b) Photoelectric current versus potential difference graphs for varying frequency (keeping same metal plate and same intensity of incident light) : These graphs indicate that the stopping potential is constant for a given frequency. The stopping potential increases with increase of frequency. The KE of the emitted electrons is proportional to the frequency of incident light. Stopping potential B1 B2 B3 ν0 A1 A2 A3 Frequency (c) Stopping potential versus frequency graphs for different metals : These graphs indicate that the stops is same for all metal, since they are parallel straight lines. The slope is a universal constant (=h/e). Further, the threshold frequency varies with emitter since the intercepts on frequency axis are different for different metals. Solved Examples 1. (i) A stopping potential of 0.82 V is required to stop the emission of photoelectrons from the surface of a metal by light of wavelength 4000 Å. For light of wavelength 3000 Å, the stopping potential is 1.85 V. Find the value of Planck's constant. (ii) At stopping potential, if the wavelength of the incident light is kept at 4000 Å but the intensity of light is increased two times, will photoelectric current be obtained? Give reasons for your answer. Sol. (i) We have hc = eV1 + W λ1 XtraEdge for IIT-JEE 23 FEBRUARY 2012 and ⇒ or h = hc = eV2 + W λ2 ⎛ 1 1 ⎞ hc ⎜ − ⎟ = e(V2 – V1) ⎝ λ2 λ1 ⎠ −19 e( V2 − V1) 1. 6× 10 ( 1. 85 − 0. 82) = ⎛ 1 1 ⎞ 8⎛ 1 1 ⎞ e ⎜ − ⎟ 3× 10 ⎜ − ⎟ ⎝ λ2 λ −7 −7 1 ⎠ ⎝ 3× 10 4× 10 ⎠ = 6.592 × 10 –34 Js (ii) No, because the stopping potential depends only on the wavelength of light and not on its intensity. 2. A small plate of a metal (work function = 1.17 eV) is plated at a distance of 2m from a monochromatic light source of wavelength 4.8 × 10 –7 m and power 1.0 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per square metre per second. If a constant magnetic field of strength 10 –4 tesla is parallel to the metal surface, find the radius of the largest circular path followed by the emitted photoelectrons. hc Sol. Energy of one photon = = λ 6. 6× 10 −34 4. 8× 10 × 3× 10 −7 = 4.125 × 10 –19 J Number of photons emitted per second 1. 0 = = 2.424 × 10 −19 4. 125× 10 18 Number of photons striking the plate per square metre per second 18 2. 424× 10 = = 4.82 × 10 2 4× 3. 14× ( 2) 16 Maximum kinetic energy of photoelectrons emitted from the plate hc Emax = – W λ = 4.125 × 10 –19 – 1.17 × 1.6 × 10 –19 = 2.253 × 10 –19 J 8
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The intercept of V0 versus ν graph on frequency<br />
axis is equal to threshold frequency (ν0). From<br />
this, the work function (hν0) can be calculated.<br />
Graphs in photoelectric effect :<br />
(a) Photoelectric current versus potential difference<br />
graphs for varying intensity (keeping same metal<br />
plate and same frequency of incident light) :<br />
These graphs indicate that stopping potential is<br />
independent of the intensity and saturation current<br />
is directly proportional to the intensity of light.<br />
ν2>ν1<br />
ν2<br />
ν1<br />
I<br />
– (V0) 2 (V0) 1<br />
Potential difference<br />
+<br />
(b) Photoelectric current versus potential difference<br />
graphs for varying frequency (keeping same<br />
metal plate and same intensity of incident light) :<br />
These graphs indicate that the stopping potential<br />
is constant for a given frequency. The stopping<br />
potential increases with increase of frequency.<br />
The KE of the emitted electrons is proportional to<br />
the frequency of incident light.<br />
Stopping potential<br />
B1 B2 B3 ν0<br />
A1 A2 A3<br />
Frequency<br />
(c) Stopping potential versus frequency graphs for<br />
different metals : These graphs indicate that the<br />
stops is same for all metal, since they are parallel<br />
straight lines. The slope is a universal constant<br />
(=h/e). Further, the threshold frequency varies<br />
with emitter since the intercepts on frequency axis<br />
are different for different metals.<br />
Solved Examples<br />
<strong>1.</strong> (i) A stopping potential of 0.82 V is required to stop<br />
the emission of photoelectrons from the surface<br />
of a metal by light of wavelength 4000 Å. For<br />
light of wavelength 3000 Å, the stopping<br />
potential is <strong>1.</strong>85 V. Find the value of Planck's<br />
constant.<br />
(ii) At stopping potential, if the wavelength of the<br />
incident light is kept at 4000 Å but the intensity<br />
of light is increased two times, will photoelectric<br />
current be obtained? Give reasons for your<br />
answer.<br />
Sol. (i) We have<br />
hc<br />
= eV1 + W<br />
λ1<br />
<strong>Xtra</strong><strong>Edge</strong> for IIT-JEE 23 FEBRUARY <strong>2012</strong><br />
and<br />
⇒<br />
or h =<br />
hc<br />
= eV2 + W<br />
λ2<br />
⎛ 1 1 ⎞<br />
hc<br />
⎜ −<br />
⎟ = e(V2 – V1)<br />
⎝ λ2 λ1<br />
⎠<br />
−19<br />
e(<br />
V2<br />
− V1)<br />
<strong>1.</strong><br />
6×<br />
10 ( <strong>1.</strong><br />
85 − 0.<br />
82)<br />
=<br />
⎛ 1 1 ⎞ 8⎛<br />
1 1 ⎞<br />
e<br />
⎜ −<br />
⎟ 3×<br />
10 ⎜ − ⎟<br />
⎝ λ2<br />
λ<br />
−7<br />
−7<br />
1 ⎠ ⎝ 3×<br />
10 4×<br />
10 ⎠<br />
= 6.592 × 10 –34 Js<br />
(ii) No, because the stopping potential depends only<br />
on the wavelength of light and not on its intensity.<br />
2. A small plate of a metal (work function = <strong>1.</strong>17 eV) is<br />
plated at a distance of 2m from a monochromatic<br />
light source of wavelength 4.8 × 10 –7 m and power<br />
<strong>1.</strong>0 watt. The light falls normally on the plate. Find<br />
the number of photons striking the metal plate per<br />
square metre per second. If a constant magnetic field<br />
of strength 10 –4 tesla is parallel to the metal surface,<br />
find the radius of the largest circular path followed by<br />
the emitted photoelectrons.<br />
hc<br />
Sol. Energy of one photon = =<br />
λ<br />
6.<br />
6×<br />
10<br />
−34<br />
4.<br />
8×<br />
10<br />
× 3×<br />
10<br />
−7<br />
= 4.125 × 10 –19 J<br />
Number of photons emitted per second<br />
<strong>1.</strong><br />
0<br />
=<br />
= 2.424 × 10<br />
−19<br />
4.<br />
125×<br />
10<br />
18<br />
Number of photons striking the plate per square<br />
metre per second<br />
18<br />
2.<br />
424×<br />
10<br />
=<br />
= 4.82 × 10<br />
2<br />
4×<br />
3.<br />
14×<br />
( 2)<br />
16<br />
Maximum kinetic energy of photoelectrons emitted<br />
from the plate<br />
hc<br />
Emax = – W<br />
λ<br />
= 4.125 × 10 –19 – <strong>1.</strong>17 × <strong>1.</strong>6 × 10 –19<br />
= 2.253 × 10 –19 J<br />
8