The document discusses the dual nature of matter and radiation. It provides answers to multiple choice and numerical questions related to photoelectric effect, de Broglie wavelength, and magnetic effect of current. Regarding photoelectric effect, it explains that electron emission from a zinc plate in ultraviolet light is due to the photoelectric effect. It also discusses how kinetic energy of photoelectrons varies with frequency of incident radiation. Regarding magnetic effect of current, it describes how to determine direction and magnitude of magnetic field around current carrying wires using the right hand grip rule. It also solves problems related to forces experienced by charged particles in magnetic fields.

1. 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?
Answer
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.
I
O V
Answer
Frequency of incident radiation was kept constant.
3. Are matter waves electromagnetic?
Answer
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?
λ
Answer
pn
We have λ= h/p = h.p-1
→ λ∞ p-1
n= -1
5. Represent graphically the variation of the de-Broglie wavelength with linear momentum
of the particle.

2. Answer
We have λ = h/p
λ∞1/pλ
The graph between h and p is shown
p
2 Marks Questions and Answers
An electron and an alpha particle have the same de-Broglie wavelength associated with them.
How are their kinetic energies related to each other?
Answer
λelectron=λα
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.
1. 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?
Answer
(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
ejected.
2. 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.
Answer
From Einstein’s photoelectric equation
Ek = hv – hv0 V0
eV0 = hv – hv0
ν

3. Vo = (h/e).v –(h/e).v0v0 Clearly Vo – v graph is a straight line of form y = mx + c; the
slope of graph is m= h/e
3. 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?
B
λ A
1/√V
Answer
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.
4. 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?
Answer
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)
˃ 2
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.

4. 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
decreased?
Answer
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
frequency.
Surface A: As no photoemission takes place; energy of incident photon is less than the
work function.
In other words, the frequency v = 1015 Hz of incident light is less than the threshold
frequency (v0)
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)?
Answer
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 ?
Answer
Let the total number of photons in the beam/sec be n
Then nhv = 39.8
Or nhc/λ = 39.8

5. 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
of sight?
Answer
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?
Answer
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.

6. 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 R.
Ans:
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
=
4𝜋𝑋10−7
2𝜋
(20/0.10 -30/0.3)
= 2 X 10-5T 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?
Ans:
B1= 𝜇0I1/2𝜋r1=
4𝜋𝑋10−7𝑋2
2𝜋𝑋4𝑋10 −2
=10-5T.
Normally into the plane of the paper.
B2= 𝜇0I2/2𝜋r2=
4𝜋𝑋10−7𝑋1.5
2𝜋𝑋3𝑋10 −2
=10-5T
normally into the plane of paper
B= B1+B2= 2 X 10-5T.
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?

7. 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
𝜇0 𝐼1(0.03)2
2[(0.03)2 + (0.04)2]
3
2
=
𝜇0 𝐼2(0.04)2
2[(0.03)2 + (0.04)2]
3
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
B =
𝜇0 𝐼
4𝑟
𝐵 =
4𝜋𝑋10−7
𝑋12
4 𝑋 0.02
= 1.9 𝑋 10−4
𝑇
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 T
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.̂

8. 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?
Ans:
B = 𝜇0I/2𝜋d.
F = qvB= q X V X 𝜇0I/2𝜋d.
F = 𝜇0Iqv/2𝜋d
Direction towards right.
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