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CHHANDAK SIR PHYSICS CLASSES
Full Marks- 20
1.a) Explain how Plank’s constant can be determined from photoelectric effect. [3]
b) An electron gun with its anode at a potential of 100V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (10 -2 mm of Hg). A magnetic field of 2.80x10
-4 T curves the path of the electrons in a circular path of radius 12.0cm. Determine e/m from the data. [2]
2.a) At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 seconds, the number of undecayed nuclei reduces to 12.5%. Calculate:
(i) Mean-life of the nuclei, (ii) the time in which the number of decayed nuclei will further reduce to 6.25% of the reduced number. [3]
b) Derive N =N
0
(
1
2
) n [2]
3a). Drawing a labeled circuit diagram, explain how p-n-p transistor can be used as an amplifier in C-E
configuration. [3]
b) Explain LED [2]
4. a) Wavelength of the first (H
α
) of Balmer series of hydrogen is 656.3mm. Find the wavelength of its
2 nd line (H
β
). [3]
b)(i) State Moseley’s law.
(ii) Estimate the value of the wavelength of K
α
–line of silver (Z=47). Given R= 109737 cm -1 . [2]
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1.Answer:
(a) Millikan performed an experiment to verify Einstein’s photoelectric equation and to determine the value of
Plank’s constant h. For, this he measured the maximum energies of photoelectrons emitted by a number of alkali metals over a wide range of incident light frequencies. His apparatus is shown in the fig.
The different alkali metal surface is mounted on a drum D placed inside an evacuated chamber. The drum can be rotated from outside so that any surface can be scraped clean by a sharp knife K and then brought before a window W. A beam of monochromatic light passing through the window falls on the surface C. The emitted photoelectrons are collected by an electrode A by means of a potential difference applied between C and A through a potential divider. The galvanometer G measures this photoelectric current.
W
K
D
C A
G
When the electrode A is given a negative potential with respect to C, the photoelectrons emitted from C are retarted.
If the potential is small, only the slower photoelectrons are pulled back and the current falls. As the negative potential is increased, more and more faster photoelectrons are pulled back. Ultimately, at a certain potential V
0 even the fastest photoelectrons fail to reach A. The photoelectric current is then zero. The (negative) potential V
0
at which the current is just reduced to zero is called the ‘stopping potential’. Since, the energy of the fastest photoelectrons is just annulled when they fall through the stopping potential , we have:
(
1
2 mv 2 ) max
=eV
0
……………………….(1)
Where e is the electronic charge and V
0
is the stopping potential. Thus the stopping potential multiplied by the electronic charge gives the maximum kinetic energy of the photoelectrons.
Millikan plotted graph between the stopping potential V
0
and the frequency v of light over a wide range of frequencies, and obtained a straight line. Parallel lines were obtained for other metallic surfaces.
Stopping potential V
0
Slope = ℎ 𝑒 𝑣
0
− ℎ𝑣
0
Frequency v 𝑒
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Now, the Einstein’s photoelectric equation is, (
1
2
Substituting the value of (
1
2 mv 2 ) mv 2 ) max
= h( v-v
0 max
from equation (1), we have ,
).
eV
0
= h( v-v
0
)
or, V
0
= ( ℎ 𝑒
) v – ( ℎ 𝑒
) v
0
……………………….(2) since h and e are constant for a given surface, equation (2) indicates that the graph between the Stopping potential
V
0 and the frequency v must be a straight line. This is actually case, as found by Millikan. Hence Einstein’s equation stands verified.
Now equation (2) shows that the slop of the straight line graph is h/e, and its intercepts on the potential axis is
− ℎ𝑣
0
. Hence, by measuring the slop of the straight li8ne obtained and using the known value of e, the value o h 𝑒 can also be obtained.
(b) Energy acquired by electrons = eV
K.E of electron,
1
2 mv 2 =eV
∴ v 2 =
2𝑒𝑉 𝑚 𝑚𝑣
2
But, evB = 𝑟
∴ v=
Or, v 2 𝑒𝐵𝑟 𝑚 𝑒𝐵𝑟
= ( 𝑚
) 2
Equating (1) and (2),
2𝑒𝑉 𝑚 𝑒 𝑚 𝑒 𝑚
=
=
=
𝐵 𝑒
2
𝐵
2𝑉 𝑚
2
2 𝑟
2
2 𝑟
2
2 𝑥 100
2.80 𝑥 10
−4
𝑥 12 𝑥 10
−2
= 1.77 x 10 11 C/kg.
2.Answer:
a)(i) The number of undecayed nuclei reduces from 25% to 12.5%, i.e becomes halved in 10 sec. It means that the half life of the nuclei is 10 sec. , that is
T = 10sec But T =
0.6931
where λ is decay constant.
Then, λ=
0.6931
𝑇
= 𝜆
0.6931
10
Hence, the mean life of the sample is τ=
1 𝜆
=
10𝑠𝑒𝑐
0.6931
= 14.435 sec.
(ii) The 6.25% of the reduced number (12.5%) is
6.25
100
x 12.5% = 0.78125%
Suppose, after 10 sec, the number of decayed nuclei reduces from 12.5% to 0.78125% in an interval of n halflives. Then, (
1
2
) n
∴ n = 4
=
0.781258
125
or 2 n
12.5
=
0.78125
= 16 =2 4
(b) Suppose, initially (at time t=0) the number of atoms in a substance is N
0
and the half-life of the substance is T.
Then , after one half-life, the number of atoms left in the substance will be N
0
/2, that is, the number of atoms at time t = T is
N =
𝑁
0
2
= N
0
(
1
2
) 1
After two half-lives, the number of atom left will be N
0
/4, that is , at time t= 2T, the number of atoms left is,
N =
𝑁
0
4
= N
0
(
1
2
) 2
Again after three half-lives, N
N =
𝑁
0
8
= N
0
(
1
2
0
)
/8 atom left, is , at time t= 3T, the number of atoms left is,
3
Thus, the number of atoms left after n half-lives is given by,
is , at time t= 2t, the number of atoms left is,
N = N
0
(
1
2
) n
3.Answer:
(a) The input circuit is forward biased by a low voltage battery V
BE
so that the resistance of the input circuit is small. The output circuit is reverse biased by means of a high-voltage battery V
CC
so that the resistance of the output circuit is high. R
L
is a load resistance connected in the collector emitted output circuit. The weak input a-c signal is applied across the base emitter circuit and the amplified output signal is obtained across the collector emitter circuit.
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I
E
= i
B
+i
C
C i
C
I
B
B
E
R
L
i
E
V
CE
i
C
input signal V
CC
output signal
I
B
i
C
V
BE
(b) It is forward biased p-n junction which emits light spontaneously. When a p-n junction diode is forward biased both the electrons and the holes move towards the junction. As they cross the junction, the electrons fall into the holes(recombine), and energy is released at the junction in the form of electromagnetic radiation. This is because the electrons fall from a higher level in the conduction band to a lower level in the valance band.
R
Light
15.(a) The wavelength formula for hydrogen atom is
1 𝜆
= R(
1 𝑛
2
1
−
1 𝑛
2
2
)
The wavelength of 1 st line (H
α
) of Balmer series is,
The wavelength of 2 nd line (H
β
) of Balmer series is,
1
= R(
2
1 𝜆
𝐻𝛼
1 𝜆
𝐻𝛽
1
= R(
2
2
2
−
−
1
3
2
) =
5𝑅
36
1
4
2
) =
3𝑅
16
Dividing the two equations, 𝜆
𝐻𝛽 𝜆
𝐻𝛼
5
=
36
𝑥
16
=
3
20
27 𝜆
𝐻 𝛽
=
20
27
𝑥 𝜆
𝐻 𝛼
=
20
27
x 656.3mm = 486.15mm
(b) (i) The squre- root of the frequency of a K-line is closely proportional to the atomic number of the emitting element. This is called Moseley’s law and may be expressed as
√𝑉 = k(Z-b)
Where Z is the atomic number of the element, and k and b are constants for a given transition of K-series.
(ii) From Moseley,s law , the wavelength of K
α
- line is given by,
1 𝜆
= (Z-1) 2 (
1
1
2
−
1
2
2
) =
3
4
3
(Z-1) 2 R
=
4
x (46) 2 x109737 cm -1
= 1.74x10
8 cm -1
∴ λ =
1
1.74 𝑥 10
8 cm = 0.57Å
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