Assignment Solutions- Dual Nature

Assignment
SolutionsDual Nature
September 19
2013
CH 4 – DUAL NATURE OF RADIATION & MATTER
SOLUTIONS
No.
Constants used: , h = 6.625 x 10-34 Js, e = 1.6 x 10-19C, c = 3 x 108 m/s
Answers
Two metals A, B have work functions 2 eV and 4 eV respectively. Which metal has a
lower threshold wavelength for photo electric effect?
1
Given βˆ…A < βˆ…B
𝒉𝝂𝒐 = βˆ… i.e.
𝑲
𝝀𝒐
= βˆ…
𝒉π‘ͺ
𝝀𝒐
= βˆ…
B
𝑲- constant
Since βˆ…A < βˆ…B , 𝝀𝑩 < 𝝀𝑨
The de Broglie wave length associated with an electron accelerated through a potential
difference V is Ξ». What will be its wavelength, when the acceleration potential is
increased to 4V?
2
For an electron,
𝝀𝒆 =
𝟏𝟐.πŸπŸ•
𝑽
For a potential 4V, 𝝀𝒆′ ∝
Å
i.e.
𝟏
πŸ’π‘½
𝟏
𝑽
𝝀𝒆 ∝
𝝀𝒆′ =
Ξ»/2
𝝀𝒆
𝟐
A particle of a mass M at rest decays into 2 particles of masses m1 and m2 having nonzero velocities. What is the ratio of the de-Broglie wavelength of the two particles?
3
Using the conservation of momentum , 𝒑1+ 𝒑2= 0
1:1
i.e. 𝒑1 = 𝒑2
we know that , 𝝀 =
𝒉
𝒑
; since 𝒑1 = 𝒑2 ;
𝝀1 = 𝝀 2
An electron, alpha particle and a proton have the same K.E. Which of these particles has
the shortest de Broglie wavelength?
Given : 𝑬𝒆 = π‘¬πœΆ = 𝑬𝒑 ; also π’Žπ’† < π‘šπ’‘ < π‘šπœΆ
Kinetic energy 𝑬 = ½ π’Ž 𝒗 2 ; 𝑬 =
∴
𝒑=
π’ŽπŸ 𝐯 𝟐
πŸπ’Ž
=
π’‘πŸ
πŸπ’Ž
𝟐 π’Ž 𝑬 ------- (1)
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de-Broglie’s relation gives , 𝝀 =
𝒉
𝒑
=
Ξ±
𝒉
πŸπ’Žπ‘¬
-------(2)
Since E is same for all the particles, (2) can be written as :
𝝀 =
π’Œ
π’Ž
Since mass of 𝜢 particle (π’ŽπœΆ) is greatest, it has the shortest de-Broglie wavelength.
2
The threshold frequency of metal is 0. When the light of frequency 20 is incident on
the metal plate, the maximum velocity of electrons emitted is v1. When the frequency of
the incident radiation is increased to 50, the maximum velocity of electrons emitted is
v2. Find the ratio of v1 to v2.
From Einstein’s photo-electric equation, 𝒉𝝂 = βˆ…o + ½ π’Ž 𝒗 2
For the two cases,
5
πŸπ’‰π‚π’ = 𝒉𝝂𝒐 + ½ π’Ž 𝒗 𝟐𝟏
𝒉𝝂𝒐 = ½ π’Ž 𝒗 𝟐𝟏 -------------------(3)
πŸ’π’‰π‚π’ = ½ π’Ž 𝒗 𝟐𝟐 ------------------(4)
Similarly,
(4) / (3), we get,
4=
𝒗 𝟐𝟐
1:2
∴ π’—πŸ : π’—πŸ = 1 : 2
𝒗 𝟐𝟏
An alpha particle and a proton are accelerated thru the same potential difference.
Calculate the ratio of linear momentum acquired by the two.
We know that, π’Žp < π‘šΞ± ; but π’ŽΞ± = πŸ’π’Žp -----(5) and 𝒒α= πŸπ’’p -----(6)
When a charge q is accelerated through a potential V, its kinetic energy E is given by :
6
E = qV -------(7)
∴ From (1) the momentum of the particle is 𝒑 =
∴ π’‘πœΆ =
𝟐 π’ŽπœΆ π’’πœΆ 𝑽
𝒑𝒑 =
𝟐 π’Ž 𝒒𝑽
𝟐 𝟐:1
𝟐 π’Žπ’‘ 𝒒𝒑 𝑽
From (5) and (6), we get, π’‘πœΆ : 𝒑𝒑 = 𝟐 𝟐 : 1
When the light of wavelength 400nm is incident on the cathode of a photocell, the
stopping potential recorded is 6 V. If the wave of incident light is increased to 600nm.
Calculate the new stopping potential.
From Einstein’s photo-electric equation, 𝒉𝝂 = βˆ…o + 𝒆𝑽𝒐 --------------(8)
∴
𝒉π‘ͺ
𝝀
= βˆ…o + 𝒆𝑽𝒐 -------------(9)
For the two cases,
𝒉π‘ͺ
𝝀
7
= βˆ…o + 𝒆𝑽𝒐 --------------(10) and
𝒉π‘ͺ
𝝀′
= βˆ…o + 𝒆𝑽′𝒐 -----------(11)
(11) – (10), we get,
𝒉𝒄 𝟏
𝒆 𝝀′
βˆ’
𝟏
𝝀
=
𝑽′𝒐 βˆ’ 𝑽𝒐
Substituting the values of h, c, e, Ξ» = 400nm, λ’ = 600nm and 𝑽𝒐 = 6V, 𝑽′𝒐 = 4.96 V
3
4.96 V
8
9
10
If the intensity of radiation incident on a photocell is doubled, what happens to the
number of photoelectrons and energy of the photoelectrons ?
As intensity of radiation is increased, the number of photoelectrons increases, but
energy of the photons remains the same. Energy will increase only with increase in
frequency of radiation, not with increase in intensity.
If the frequency of radiation incident on a photocell is doubled, what happens to the
number of photoelectrons and energy of the photoelectrons ?
Same as above.
Light of two different frequencies whose photons have energies 1eV and 2.5eV,
respectively, successively illuminate a metal whose work function is 0.5 eV. Find the
ratio of maximum speed of the emitted electrons.
𝒉𝝂𝒐 = βˆ…o + ½ π’Ž 𝒗 𝟐𝟏
1eV = 0.5 ev + ½ π’Ž 𝒗 𝟐𝟏
0.5 eV = ½ π’Ž 𝒗 𝟐𝟏 ------------------(1)
2.5eV = 0.5 ev + ½ π’Ž 𝒗 𝟐𝟐
2eV = ½ π’Ž 𝒗 𝟐𝟐 --------------------(2)
(1) / (2) we get,
𝟏
πŸ’
=
𝒗 𝟐𝟏
𝒗 𝟐𝟐
Increases
No Change
No Change
Increases
1:2
∴ 𝒗 𝟏 : 𝒗 𝟐 = 1:2
A proton when accelerated through a potential difference of V volt has a de-Broglie
wavelength Ξ». What should be the potential required to accelerate an Ξ± particle in order
to have the same Ξ» ?
Given λα = λp ,
The de-Broglie wavelength of a charge accelerated through a potential V is given by:
𝝀=
11
𝒉
𝟐 π’Ž 𝒒𝑽
𝝀𝒑 =
𝒉
----(12) and
𝟐 π’Žπ’‘ 𝒒𝒑 𝑽
π€πœΆ =
𝒉
𝟐 π’ŽπœΆ π’’πœΆ 𝑽′
---(13)
V/8 volt
But π’ŽΞ± = πŸ’π’Žp, 𝒒α= πŸπ’’p and λα = Ξ»p (given)
∴ equating (12) and (13), we get,
𝑽
12
𝑽′ = πŸ–
When light of wavelength 0.6mm falls on a photocell photoelectrons are emitted for
which the stopping potential is 0.5V. With light of wavelength 0.04 mm the stopping
potential changes to 1.5V. Find the work-function of the metal in electron-volts.
If 5% of energy supplied to a bulb is irradiated as visible light, how many quanta are
emitted per second by a 100W bulb ? Assume wavelength of light is 5.6 x 10-5 cm.
13
100W = 100 J/s
Only 5% is visible radiation. Ie 5 J/s
π‘ͺ
Energy of one photon, 𝑬 = 𝒉𝝂 or 𝑬 = 𝒉 𝝀 ----------(14)
Energy of β€˜n’ photons = 5 J
βˆ΄π’=
πŸ“
π‘ͺ
𝒉
𝝀
= 1.4 x 1019 s-1
4
1.4 x 1019 s-1
What is the ratio of the wavelength of a photon and that of an electron of the same
energy ?
π‘ͺ
14
π‘ͺ
For a photon, 𝑬 = 𝒉 𝝀
𝝀𝒑
𝝀𝒆
=𝒄
πŸπ’Žπ‘¬
𝑬
= 𝒄
i.e. 𝝀 𝒑 = 𝒉 𝑬 ---(15) For an electron, 𝝀 𝒆 =
𝒉
𝒑
=
𝒉
πŸπ’Žπ‘¬
---(16)
πŸπ’Ž
𝒄
πŸπ’Ž
𝑬
𝑬
The graph below shows the variation of the max. kinetic energy of photoelectrons with
frequency of incident light. Find the threshold frequency and work function of the metal
from the graph.
E (eV)
8
0
10
20
Ξ½ (x 1014 Hz)
30
1015 Hz
4 eV
-4
15
From Einstein’s photoelectric equation, 𝒉𝝂 = βˆ…o + 𝒆𝑽 -----(17)
∴ 𝒆𝑽 = 𝒉𝝂 - βˆ…o ----(18) (equation for a straight line with slope = h)
But 𝒉𝝂𝒐 = βˆ…o
i.e. when 𝝂 = 𝝂𝒐 , 𝒆𝑽 = 𝟎. From the graph 𝝂𝒐 = 1015 Hz
From the graph and from (18) βˆ…π’ = πŸ’π’†π‘½
Eby P Kurien, [email protected]
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