Dual Nature of Matter and Radiation
1 Mark Questions and Answers
1. An uncharged zinc plate becomes positively charged when irradiated by ultraviolet
radiations. What is the phenomenon due to?
The phenomenon of electron emission is called photoelectric effect.
2. In an experiment on photoelectric effect, the photoelectric current (I) and the anode,
the following graphs were obtained between potential (V). Name the characteristic of
the incident radiation that was kept constant in this experiment.
Frequency of incident radiation was kept constant.
3. Are matter waves electromagnetic?
No, Matter waves are not electromagnetic as they are associated with neutral particles.
They are probability waves which tell the probability of location of particle in a certain
region of space.
4. A student plots graph between de-Broglie wavelength and some power (n) of
momentum. What is the value of n?
We have λ= h/p = h.p-1
→ λ∞ p-1
5. Represent graphically the variation of the de-Broglie wavelength with linear momentum
of the particle.
We have λ = h/p
The graph between h and p is shown
2 Marks Questions and Answers
1. An electron and an alpha particle have the same de-Broglie wavelength associated with
them. How are their kinetic energies related to each other?
de-Broglie wavelength associated with a particle of mass m and energy E is
λ = h/√2mE
h/√2meEe = h/√2maEa
That is kinetic energy of electron and a particle are in inverse ratio of these masses.
2. Electrons are emitted from the surface when green light is incident on it, but no
electrons are ejected when yellow light is incident on it. Do you expect electrons to be
ejected when surface is exposed to (i) red light (ii) blue light?
(i) The wavelength of red light is longer than threshold wavelength hence no
electron will be emitted with red lights.
(ii) The wavelength of blue light is smaller than threshold wavelength hence will be
3. Draw a graph showing the variation of stopping potential with frequency of incident
radiation in relation to photoelectric effect. Deduce an expression for the slope of this
graph using Einstein’s photoelectric equation.
From Einstein’s photoelectric equation
Ek = hv – hv0 V0
eV0 = hv – hv0
Vo = (h/e).v –(h/e).v0v0 ν
Clearly Vo – v graph is a straight line of form y = mx + c; the slope of graph is m = h/e
4. Two lines A and B shown in the graph represent the de-Broglie wavelength (λ) as a
function of 1/√V (V is the accelerating potential) for two particles having the same
charge. Which of the two represents the particle of smaller mass?
De-Broglie wavelengthλ = h/√2mqV or λ = h/√2mq . 1/√V
The graph of λ versus 1/√V is a straight line of slope h/√2mq α 1/√m. the slope of line B
is large. So particle B has smaller mass.
5. If the frequency of light falling on a metal is doubled, what will be the effect on photo
current and the maximum kinetic energy of emitted photoelectrons?
The photocurrent does not depend on the frequency of incident radiation, hence the
photocurrent remains unchanged. The maximum kinetic energy increases with increase
of frequency, given by EK = hν – W
If frequency is doubled, EK’ = 2 hν– W
EK’ / EK= ( 2 hν – 2W +W) / (hν – W)
= 2 + W / (hν – W)
i.e maximum kinetic energy will increase to slightly more than double value.
3 Mark Questions and Answers:
1. Radiations of frequency 1015 Hz are incident on two photosensitive surfaces A and B.
following observations are recorded.
Surface A: No Photoemission takes place.
Surface B: Photoemission takes place but photoelectrons have zero energy.
Explain the above observations on the basis of Einstein’s photoelectric equation. How
will the observation with surface B change when wavelength of incident light is
Einstein’s Photoelectric equation is
Hν = W + Ek→Ek= hν – W orEk = hν – hν0
Where W is work function of metal, ν is frequency of incident light and ν0 is threshold
Surface A: As no photoemission takes place; energy of incident photon is less than the
In other words, the frequency v = 1015 Hz of incident light is less than the threshold
Surface B: As phenomenon takes place with zero kinetic energy of photoelectrons
(i.e, Ek = 0), then equation (1) gives W = hv or v0 = v.
i.e., energy of incident photon is equal to work function. In other words, threshold
frequency of metal is equal to frequency of incident photon i.e., v0 = 1015 Hz. When
wavelength of incident light is decreased, the energy of incident photon becomes more
than the work function, so photoelectrons emitted will have finite kinetic energy given
by Ek = (hc/λ )– W.
2. In a photoelectric experiment, an incident radiation of wavelength λ emits photo
electrons with maximum kinetic energy E. Show that to get the electrons with energy
2E; the wavelength of the incident radiations should be hcλ/(Eλ + hc)?
We have E = hv – φ0 = (hc/λ ) - φ0
For electron with energy 2E, let the wavelength be λ1
Then 2E = (hc/λ1) - φ0
or(2hc/λ)- 2φ0 = (hc/λ1) - φ0
hc/λ1 = (2hc/λ) - φ0
= 2hc/λ + E – hc/λ
= hc/λ + E = (hc + Eλ)/λ
λ1= hcλ/(Eλ + hc)
3. Assuming that 1 % of the photon in a 39.8 Wm-2 intensity beam of wavelength 5000 Å
cause emission of photoelectrons, calculate the resulting photoelectric current ?
Let the total number of photons in the beam/sec be n
Then nhv = 39.8
Or nhc/λ = 39.8
N = 39.8 x λ/hc = 39.8 x 5000 x 1010 / (6.63 x 10-34 x 3 x 108) = (199/19.89) x 1019 = 1020
No of photoelectrons emitted/sec = no of photons absorbed
= 1 % of 1020 = 1018
Current = 1018e = 1018 x 1.6 x 10-19
= 0.16 A = 160mA.
4. The human eye needs a minimum intensity of 10-10 Wm-2 to produce sensation of sight.
If the area of the pupil is 10-6 m2 and the wavelength of incident light is 5600 Å; what
should be the minimum number of photons entering the pupil / sec to excite the sense
Minimum intensity detectable = 10-10 W/m2
Minimum energy required on pupil
= 10-10 W/m2 x 10-6 m2
= 10-16 joule/sec
Let the number of photons entering the eye be n/sec
Then n.hv = 10-16
n = 10-16/(6.63 x 10-34 x c/λ)
= 10-16 x 5600 x 10-10/(6.63 x 10-34 x 3 x 108) = 5600/19.89 = 283
5. An electron moving with velocity v and a photon have same de-Broglie wavelength.
What will be the ratio of their energy values?
We have λe =λph
h/Pe = h/Pph→Pe = Pph
ormv = Eph/c (1)
Now 1/2mv2 = Ee→ m = 2Ee/v2
From (1) 2Ee/v = Eph/c
Hence Ee/Eph = v/2c.
Magnetic effect of current
1. Figure shows two current carrying curves 1 and 2. Find the
magnitudes and directions of the magnetic field at points p, Q and
According to right hand gap rule, B1 of wire 1 at point p will point
normally outward while B2 of wire 2 will point normally inward.
BP = B1- B2 =𝜇0I1/2𝜋r1-𝜇0I2/2𝜋r2
= 2 X 10-5
T pointing normally outward.
2. Two infinitely long insulated wires are kept perpendicular to each
other. They carry currents I1= 2A and I2=1.5 A. (i) Find the
magnitude and direction of the magnetic field at p. (ii) If the wires
, what would reserved in one of the wires, what would be the
magnitude of the field B?
Normally into the plane of the paper.
normally into the plane of paper
B= B1+B2= 2 X 10-5
3. Two co-axial circular loops L 1 and L2 of radii 3cm and 4cm are
placed as shown. What should be the magnetic and direction of
the current in the loop L2 so that the net magnetic field at the
point o be zero?
For the net magnetic field at the point ‘o’ to be zero, the direction
of current in loop L2 should be opposite to that in loop L1.
Magnitude of magnetic = Magnitude of magnetic
field due to current I1 in L1 field due to current I2 in L2
2[(0.03)2 + (0.04)2]
2[(0.03)2 + (0.04)2]
I2 = 9/16 X 1 = 0.56 A
4. A straight wire carrying a current of 12A is bent into a semicircular
are of radius 2 cm as shown in figure. What is the direction and
magnitude of 𝐵⃗ at the centre of the arc? Would your answer
change if the wire were bent into a semicircular arc of the same
radius bent in the opposite way as shown in the figure?
(i) B at the centre of the arc is
4 𝑋 0.02
= 1.9 𝑋 10−4
According to right hand rule, the direction of the field is normally
into the plane of paper.
ii) The magnetic field will be of the same magnitude
B = 1.9 X 10-4
The direction of the field is normally out of the plane of paper.
5. A beam of electron projected along +x –axis experiences a force
due to a magnetic field along the +y- axis. What is the direction of
the magnetic field.
V along X-axis- 𝑖.̂
B along y- axis-𝑗.̂
F = q ( V X B).
V X B is along 𝑖̂𝑋 𝑗̂ = 𝑘.̂
i.e. +Z – axis.
6. A long straight wire AB carries a current I. A proton p travels with
a speed V1 parallel to the wire, at a distance d from A in a
direction opposite to the current as shown in the figure. What is
the force experienced by the proton and what is its direction?
B = 𝜇0I/2𝜋d.
F = qvB= q X V X 𝜇0I/2𝜋d.
F = 𝜇0Iqv/2𝜋d
Direction towards right.
MAGNETIC EFFECT OF CURRENT
VALUE BASED QUESTIONS
1.MrNarasimhan a 65 year old person often complained of
neck pain. One
day his grandson Avinash, suggested that magnetic
therapy is very effective
in reducing such pains. He said that the permanent
used in the device will help to produce Joule’s heating
effects in the blood
stream, which helps the blood flow better.He immediately
friend in Chennai, who was running Magnetic Therapy
Narasimhan who felt better.
(a).What two values did Avinash exhibit towards his
(b). What is the SI unit of magnetic induction and define
Ans. (a)Responsiblebehaviour, concern and awareness
(b) Tesla (defn)
2. MsUdaya joined a PG course in Nanotechnology lab in
IIT Chennai. The
first day, when she went to the lab, she met Mr. Antonio,
assistant.He greeted her and advised her not to touch the
wires which were
suspended from the roof at every part of the lab as they
were from high
voltage lines. He also told her not to bring any of the two
wires closer to each
other during any experimental applications. He helped her
about the precautions that has to be taken in the lab.
(a) What value did Mr. Antonio exhibit towards Ms.
Udaya? Mention any two
(b)Why two high voltage power transmission lines should
not be close to each
(a) Responsible behaviour, sensitivity, concern for others
and alerting the
3. In the birthday party of Bharat, a class 7student, his
parents gave big
slinkys to all his friends as return gifts. The next day,
during the physics
classMr Mohan, the teacher explained them about the
magnetic fields using current carrying coil and also said
that they can make
magnets, using such coils by passing high currents
them. That night Sumanth, a friend of Bharat, asked his
father about the
coils, and their shape. His father asked him to bring the
slinky, that his
friend gave and explained the uses of toroid and solenoid.
(a)What value did Sumanth’s father exhibit towards his
(b).What is the difference in the fields produced by the
solenoid and Toroid?
Ans(a)Responsibility, makes his child to understand the
concepts and to generate interest in the subjects
b)The magnetic field lines in a toroid is concentric circles
whereas in solenoid
it is straight within the turns.
4 .MsLavanya a house wife aged 42 years complained of
stomach ache one
day. Her husband MrSrinivas took her to a nearby hospital.
observed her and found something wrong near her liver
malignancy. There after checking her MRI scan, a team of
her to go through Carbon radio therapy which is very safe.
They said using
cyclotron, high speed ions can be generated that directly
cancerous tissues and destroy them.
(a). What values did Mrsrinivas and the doctor have
exhibited? Mention any
(b).What are the role played by Electric field and magnetic
Ans.(a) concern for others, helpfulness, presence of mind,
(b) The charged particles are accelerated by the electric
magnetic field provides the necessary centripetal