Download Electron and photon with solution

Q 1: Radiation of wavelength of 180nm ejects photoelectrons from a plate whosework function is 2.0eV. If a uniform magnetic field dif flux density
5.0 × 10-5 T is applied parallel to the plate, what should be the velocity of electrons ejected normally from the plate with maximum energy?
(a) 1.3 × 106 m/s
(b) 3.0 × 106 m/s
(c) 1.3 × 105 m/s
Sol: (a) Using Einstein’s photoelectric equation
(d) 3.0 × 105 m/s
mυ2max =
(Ek)max =
eV = 2.0 X 1.6 X 10-19 J,
c = 3 X 108 m/s, h = 6.6 X 10-34 J.s
10-19 J, υmax = [
] 1/2 = 1.3 X 106 m/s
-φ
λ = 180 X 10-9 m, φ = 2.0
mυ2max = 7.8 X
On putting thesevalues, we get,
Q 2: In ques. No. 1, the radius of the path followed by the electros will be:
(a) 0.45 m
(b) 0.30 m
(c) 0.15 m
(d) 0.075 m
Beυ =
Sol: (c) Let r be the radius of circular path of electrons then
r=
=
= 0.149 ≈ 0.15 m
,
Q 3: A small plate of a metal (work function = 1.17 eV) is placed at a distanceof 2 m from a monochromatic light sourceof wavelength 4.8 × 10-7 m and
power 1.0 W. The light falls normally on the plate. Find the no. of photons striking the metal plate per squaremeter per second.
(a) 4.8 × 1015
(b) 4.8 × 1016
(c) 1.6 × 1015
(d) 1.6 × 1016
Sol: (b) Total number of photons emitted per secondfrom the light sourceN =
=
= 2.424 X 1018, Thesephotons spread out in all directions. The number of
n=
=
N=
=
photons striking on 1 m2 surfaceper secondis
= 4.825 X 1016
Q 4: In ques. No. 3, if a constant uniform magnetic field of strength 10-4 T is applied parallel to the metal surface, find the radius of the largest circular
path followed by the emitted photo electrons:
(a) 8 cm
(b) 4 cm
(c) 2 cm
(d) 1 cm
hν = φ +
Sol: (b) From Einstein’s photoelectric equation
J, φ = 1.17 X 1.6 X 10-19 J,
m = 9.1 X 10-31 kg
υmax = 7 X 105 m/s
given by
mυ2max, hc/λ = φ +
mυ2max
….(i)
In Q. no. 3
=
On purring thesevaluesin eq. (i), we get,
Radius of circular path followed by the electrons in the magnetic field B is
r=
=
,
= 4 X 10-2 m = 4 cm
Q 5: The wavelength of characteristic x- rays K – line emitted by a hydrogen like element is 0.32 Å. The wavelength of K – line emitted by the same
element is:
(a) 0.17 Å
(b) 0.27 Å
(c) 0.32 Å
(d) 0.37 Å
Sol: (b) The frequency of spectral lines when electron jumps from energy level ni to nf is given by
],
= R (z – 1)2 [
(1/ λα) = R (z – 1)2 = [
For kα - line
-
],
=
=
=
,
λβ =
X 0.32 Å= 0.27 Å
]
= (1/λ) = R (z - σ)2 [
For kβ- line
(1/ λβ)
Q 6: In Millikan’s oil drop experiment, a drop of radius r and carrying a chargeq is held stationary between the plates at a potential differenceV. Then
another drop of radius 2r will be stationary at a potential difference4V if it carries a charge:
(a) q/2
(b) 2q
(c) 4q
(d) q/4
π3 ρg = qE,
Sol: (b) Downwards weight of the drop is balancedby the upwards electrostatic force.
πr3 ρg =
electric field E = V/l where l is the separation between the plates, hence
q2 = q1 (
,
or
Here
r3 ∝ qV, (r1/r 2)3 =
,
) ( )3 = 2q
) ( )3 = q (
Q 7: The maximum velocity of photoelectrons emitted from a metal surfaceis 1.76 × 106 m/s. If the e/m ratio of electron is 1.76 × 1011 C/kg, then the
stopping potential of the metal is:
(a) 2.2 V
(b) 4.4 V
Sol: (c) eVs =
mυ2max,
(c) 8.8 V
VS =
(d) 17.6 V
=
,
= 8.8 V
Q 8: When a monochromatic sourceof light is at a distanceof 0.2 m from a photoelectric cell, the cut off voltage and the saturation current are
respectively 0.6 V and 18 mA. If the samesourceis placed0.6 m away from the cell then:
(a) The stopping potential will be 0.2 V
(b) The
stopping potential will be 1.8 V
(c) The saturation current will be 6.0 mA
(d) The saturation current will be 2.0 mA
Sol: (d) The stopping potential (or the cut-off voltage) dependson the frequency of incident light and is independent to the intensity of incident light
radiations. Hence, stopping potential still remains 0.6 V.
between sourceof light
However, the intensity of light varies with distancer as 1/r2. When the distance
and the cell is increasedfrom 0.2 m to 0.6 m, the intensity reducesfrom I to . Therefore
photoelectric current i also
reducesto i/9 i.e.., 18/9 = 2.0 mA.
Q 9: The maximum kinetic energy of photoelectron emitted from the surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential is:
(a) 2 V
(b) 4 V
(c) 6 V
Sol: (b) hν = hν0 + eVs,
(d) 10 V
where
eVS =
mυ2max = 4 eV,
or
VS = 4 V
Q 10: A photon of energy 8 eV is inclined on a metal surfaceof threshold frequency 1.6 × 1015 Hz. The kinetic energy ( in eV) of the photoelectron
emitted is:(Take h = 6 × 10-34 Js)
(a) 6
(b) 1.6
(c) 1.2
Sol: (d) hν = hν0 + (Ek)max
(d) 2
(Ek)max = hν - hν0,
= 3.2 X 10-19 J,
= 1.8 X 1.6 X 10-19 – 6 X 10-34 X 1.6 X 1015
=
= 1.6 X 10-19 X 2
= 2 eV
Q 11: The work function of a substanceis 4.0 eV. The longest wavelength of light that can causephotoelectron emission from this substanceis
approximately: (a) 540 nm
(b) 400 nm
(c) 310 nm
Sol: (c) hν0 = (hc/λmax) = 4.0 eV = 4 X 1.6 X 10-19 J,
(d) 220 nm
λmax =
=
,
= 3.1
X 10-7 m = 310 nm
Q 12: A particle of massM at rest decaysinto two articles of massesm1 and m2, having non – zero velocities. The ratio of de- Broglie wavelengths of the
particles,
(a)
m
m
1
2
/
1
2
is:
(b)
m
m
2
1
(c) 1.0
(d)
m
m
2
1
m1υ1 + m2υ2 = 0,
Sol: (c) Using the principle of conservation of linear momentum,
λ1 =
wavelengths,
λ2 =
,
=
m1υ1 = - m2υ2,
p1 = - p2 de-Broglie
= 1.0
Q 13: Photoelectric emission is observedfrom a metallic surfacefor frequenciesv1 and v2 of the incident light (v1 > v2). If the maximum valuesof kinetic
energy of the photoelectrons emitted in the two casesare in the ratio 1 : n, then the threshold frequency of the metallic surfaceis:
(a)
v −v
1
2
(b)
n −1
n v1 − v2
(c)
n −1
Sol: (b) (KE)max = E k = hν - hν0,
0
2
1
(d)
n −1
=
nν1 - nν0 = ν2 - ν0, or
Q 14: If
nv − v
,
=
ν0 (n – 1) = nν1 - ν2, or
v −v
1
2
n
or
n=
ν0 =
is the de – Broglie wavelength for a proton accelerated through a potential differenceof 100 V, the de – Broglie wavelength for
–
particle accelerated through samepotential differenceis:
(a)
2 2 λ0
(b)
λ
0
(c)
2
λ
0
(d)
2 2
λ
0
2
λ=
Sol: (c) De-Broglie wavelength associated with a particle of massm and chargeq is given by
which the particle is accelerated,
=
=
Thus,
λα =
,
λ∝
,
=
,
where V is the potential differenceto
or
= ;
=
= ,
So,
=
Q 15: Two photons of energiestwice and thrice the work function of a metal are incident on the metal surface. Then the ratio of maximum velocities of
the photoelectron emitted in the two casesrespectively is:
(a) √2 : 1
(b) √3 : 3
(c) √3 : √2
Sol: (d) In the first case,hν = 2φ,
φ+
⇒
= 2φ,
2φ = φ +
So,
=
= ,
(d) 1 : √2
⇒
= φ,
and in the second casehν = 3φ
So,
3φ =
= 1/
Q 16: The de – Broglie wavelength of a particle moving with a velocity 2.25 × 108 m/s is equal to the wavelength of a photon. The ratio of kinetic energy
of the particle to the energy of the photon is: (Velocity of light is
(d)
3 × 108 m/s)
(a)
1
8
(b)
3
8
(c)
7
8
Sol: (b) For the particle λ = h/mυ,
photon E’ = hν = hc/λ =
=
Kinetic energy of the particle E =
= mυc
mυ2
Kinetic energy of the
or
=
=
=
Q 17: Sodium and copper have work functions 2.3 eV and 4.5 eV respectively. Then the ratio of the wavelength is nearest to:
(b) 4 : 1
=
(c) 2 : 1
(d) 1 : 4
(a) 1 : 2
,
5
8
Sol: (c) Work function φ = hν0 = hc/λ0,
=
=
≈ ,
≈
or
Q 18: two identical photo cathodesreceivelight of frequenciesf 1 and f 2. If the velocities of the photoelectron (of massm) coming out are respectively v1
and v2, then:
(a)
(c)
2
2
1
2
2
2
1
2
v −v
v +v
2h
=
(
m
2h
=
(
m
1/ 2
f
−
1
f
2h

( f 1+f 2)

m

)
(b) v1 +v2 =
)
2h

(d) v −v =
( f 1−f 2)

m

2
1/ 2
f
1
+
f
2
1
2
Sol: (a) Using Einstein’s photoelectric equation
mυ2max, or
hf = hf0 +
υ12 = (2/m) h (f 2 – f 0)
and
υ22 = (2/m) h (f 2 – f 0),
υ12 - υ22 = 2h/m (f1 – f 2)
or
Q 19: According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photoelectron from a metal versusthe frequency, of the
incident radiation givesa straight line whoseslope:
(a) Dependson the nature of the metal used
(b) Dependson the intensity of radiation
(c) Dependsboth on the intensity of radiation and the metal used
(d) Is the samefor all the metals and independent to the intensity of the radiations
Sol: (d) (KE)max = hν - φ,
Comparing this equation with the standard equation for
y = mx + c,
the straight line
m = tan θ = h,
we get,
Hencethe slopeof the straight line
givesthe value of h the planck’s constant
and is independent to the intensity of incident radiations.
Q 20: A photo cell is illuminated by a small bright sourceplaced 1 m away. When the samesourceof light is placed ½ m away, the number of electrons
emitted by photocathodewould:
(a) Decreaseby a factor of 4
(b) Increaseby a factor of 4
(c) Decreaseby a factor of 2
(d) Increaseby a factor of 2
Sol: (b) Intensity of light radiations I ∝
when the distancebetween light sourceand the photocell is changed from 1 m to
falling on the photocell increasesby a factor of 4. Hencethe rate
emission from the
photocathode
m the intensity of light
of photoelectron emission from the photocathodealso increaseby a factor
also increasesby a factor of 4.
Q 21: If the kinetic energy of a free electron doubles, its de – Broglie wavelength changesby the factor:
(a)
1
2
(b) 2
(c)
Sol: (c) de-Broglie wavelength λ = h/p,
,
So,
λ2/ λ1 =
1
2
(d)
2
(or KE ‘K’ = p2/2m)λ =
where the momentum p =
=
,
or
λ∝
λ2 = λ1/
,
Q 22: The threshold frequency for a metallic surfacecorrespondsto energy of 6.2 eV and the stopping potential for a radiation incident on this surfaceis
5 V. The incident radiation lies in:
(a) Ultra violet region
(b) Infra red region
Sol: (a) Given hν0 = 6.2 eV; V S = 5V,
11.2 eV,
λ=
,
(c) Visible region
(d) x- ray region
Using Einstein’s photoelectric equation, hν = hν0 + eVS
λ=
or
,
This wavelength goesin the ultraviolet region.
Q23: The time taken by a photo electron to comeout after the photon strikes approximately:
(a) 10-4 s
(b) 10-10 s
(c) 10-16 s
(d) 10-1 s
≈ 1.1 X 10-7 m,
hc/λ = (6.2 + 5.0) eV =
≈ 1100 Å,
Sol: (b) Experimental observations provesthat the photoelectric effect is an instantaneousphenomenonin
which photoelectron comesout after
the photon strikes the metallic surfacein approximately 10-10 s.
Q 24: The anodevoltage of a photo – cell is kept fixed. The wavelength
of a light falling on the cathodeis gradually changed. The plate current I of
the photo cell vary as follows:
I
I
I
I
(a) A
(b) B
(C
(c) C
(d) D
)
Sol: (b) The photoelectric current dependson the intensity of incident light radiations and is independent to
(B
incident light radiations. However,
(A their will be no photoelectron
wavelengths higher than the ) threshold wavelength λ0. ]
the frequency (or the wavelength) of
ejected from the cathodeand henceno photo electron current for the
(D
)
)
Q25: In a Thomsonset up, E =30 V cm-1 and B =6G. Then speedof the electron that goesundeflected in the commonregion of the two fields will be:
(a) 1 × 106 m/s
(b) 5 × 106 m/s
Sol: (b) υ = E/B,
Given
G = 6 X 10-4 Wb/m2,
υ=
(c) 15 × 106 m/s
(d) 25 × 106 m/s
E = 30 V/cm = 30 x 102 V/m,
B=6
= 5 X 106 m/s
Q 26: Cathoderays are producedwhen the pressureis of the order of:
(a) 0.001 mm of Hg
(b) 0.01 mm of Hg
(c) 1 m of Hg
(d) 2 cm of Hg
(e) 0.1 cm of Hg
Sol: (b)
Q 27: Which of the following phenomenais observedwhen black dischargeoccurs in a dischargetube:
(a) The wall of the tube opposite to the cathodeglow
(b) A glow appearsat the cathode
(c) Positive column appears
(d) Striations are seen
Sol: (a)
Q 28: If the intensity of radiation incident on a photo cell be increasedfour times, then no. of photoelectrons and energy of photoelectrons emitted
become:
(a) Doubled, remains unchanged
(b) Four times, remains unchanged
(c) Remains unchanged, doubled
(d) four times, doubled
Sol: (b) The number of emitted photoelectrons is directly proportional to the intensity light. The energy of
to the intensity of incident radiations, it dependson the
emitted photoelectrons is independent
frequency of incident photons.
Q 29: When yellow light is incident on a surfaceno electrons are emitted while greenlight can emit. If red light is incident on the surfacethen:
Electrons of higher energy are emitted
(a)
(b) Electrons of lower energy are emitted (c) No electrons are emitted
(d) Photons are emitted
Sol: (c) λred > λyellow, so that νred < νyellow, Thus for photosensitive material the red colours has frequency its threshold limit. Henceno electrons are
emitted.
Q 30: Photocell is a deviceto:
(a) Store photons
(c) Convert photon energy into mechanical energy
(b) Measure light intensity
(d) Store electrical energy for replacing storagecells
Sol: (b)
Q 31: Assuming photo emission to continue to take place, the factor by which the maximum velocity of the emitted photoelectron changesapproximately
when the wavelength of the incident radiation is increasedfour times, is:
(a) 1/2
(b) 2
(c) 1/4
(d) 4
Sol: (a) mυ2max = h (ν - ν0) = hc ( -
), When λ increases4 times the velocity of the emitted photoelectrons decreasesto approximately (1/2) times.
Q 32: Energy conservation in a photoelectric cell takes placefrom:
(a) Chemical to electrical
(b) Magnetic to electrical
(c) Optical to electrical
(d) Mechanical to electrical
Sol: (c)
Q 33: The velocity of the most energetic electron emitted from a metallic surfaceis doubled when the frequency v of incident radiation is doubled. The
function of this metal is:
(a) 2/3 hv
(b) hv/3
Sol: (a) hν = φ + E k,
(c) hv/2
(d) zero
where φ = hν0 and E k =
….(i)
mυ2max,
when ν and υmax are doubled,
h (2ν) = φ + 4Ek
From eqs(i) and (ii), E k = hν/3
….(ii)
and
φ = hν - E k = hν -
,
= 2hν/3
Q 34: Light of wavelength
strikes a photosensitive surfaceand electrons are ejected with kinetic energy E. If the KE is increasedto be 2E, the
wavelength must be changeto
λ
(a)
λ
'
>
'
where:
(b) / 2 <
λ<
'
(c)
Sol: (b) Ek = E = hν - φ,
or
E=
eqs. (i) and (ii), we get,
E = hc [
- ],
(ii), we get,
2[
>
,
- φ] =
λ
'
-φ
=2
(d)
…(i)
hc [ -
'
= /2
In the second case 2E =
Since, KE is always positive
- φ, or
λ
] =φ
>
-φ
or λ’ < λ,
…(ii)
From
Further putting the value E in eq.
Since, work function φ is always positive,
λ’ > λ/2, λ/2 < λ’ < λ
Q 35: The work function of metal A and B are in the ratio 1: 2. If light of frequenciesf and 2f are incident on metal surfacesA and B respectively, the
ratio of the maximum kinetic energiesof photoelectrons emitted is: (f is greater than threshold frequency of A, 2f is greater than threshold frequency of B)
(a) 1: 1
(b) 1: 2
Sol: (b) (Ek)max = hν - φ,
for metal B,
=
(c) 1: 3
E A = hf - φA
For metal A,
E b = h (2f) - φB
…(ii),
Given that φA : φB = 1 : 1,
,
(d) 1: 4
…(i)
and
From eqns. (i) and (ii),
or
=
=
=
Q 36: Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material, photoelectric current is emitted. If the frequency of
the light is halved and intensity is doubled, the photoelectric current becomes:
(a) Half of original current
(b) 2 times the original
current
(c) 4 times the original current
(d) Zero times the original current
Sol: (d) Photoelectric current is emitted for the light of frequency ν = 1.5 ν0, where ν0 is the threshold
frequency. If frequency is halved, i.e.., ν/2 or
ν0 it becomeslessthan threshold frequency. Hence there will be no photoelectric current.
Q 37: Light of wavelength 3500 Å is incident on two metals A and B, A of work function 4.2 eV and B of work function 1.19 eV respectively. The
photoelectrons will be emitted by:
(a) Metal A
(b) Metal B
(c) Both A and B
(d) Neither metal A nor metal B
E = hν = hc/λ =
Sol: (b) The kinetic energy o incident photons is given by
=
J
= 5.69 X 10-19 J
SinceφA > E and φB < E, therefore only metal B is capable of giving photoelectric current.
eV = 3.55 eV,
Q 38: The stopping potential dependson:
(a) Intensity of incident light (b) Frequency of incident light
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Sol: (b)
Q 39: Photoelectric effect can be explained by assuming that light:
longitudinal wave
(c) Can be polarized
(a) Is a form of transversewave
(b) Is a form of
(d) Consist of quanta
Sol: (d)
Q 40: One milliwatt of light of wavelength 4560 Å is incident on a cesium surfaceof work function 1.9 eV. Given that quantum efficiency of
photoelectric emission is 0.5 %, Plank’s constant h = 6.62 × 10-34 J-s, velocity of light c = 3 × 108 m/s, the photoelectric current liberated is:
(a) 1.856 × 10-4 A
(b) 1.856 × 10-5 A
(c) 1.856 × 10-6 A
(d) 1.856 × 10-7 A
Sol: (c) The number of photons falling per secondon the cesium surface is given by (E = nhν)
(p =
= 1 mW 10-3 J/s)
n=
= 2.35 X 1015
,
Sincequantum efficiency
is 0.5%, hencethe number of photoelectrons emitted per second from the surface is, n’ = 0.5% of n =
1.175 X 1013
=
,
The photoelectric current will be
i=
=
=
=
= 1.175 X 1013 X 1.6 X 10-19 A,
= 1.856 X 10-6 A
Q 41: The maximum wave length of a beamof light that can be used to producephotoelectric effect on a metal is 250 nm. The energy of the electron in
joule emitted from the surfaceof the metal when a beamof light of wavelength 200 nm is used:
× 10-22
(a) 18.96 × 10-20
(b) 19.86 × 10-20
(c) 69.81
(d) 89.61 × 10-22
Sol: (b) Threshold wavelength λ0 = 250 nm,
= hc [ -
]
E k = hν - hν0
using Einstein’s photoelectric equation,
= 6.6 X 10-34 X 3 X 108 [
-
] X 109,
= 19.86 X 10-20 J
Q 42: Photons of 5.5 eV energy fall on the surfaceof the metal emitting photoelectrons of maximum kinetic energy 4.0 eV. The stopping voltage required
for theseelectrons is:
(a) 1.5 V
(b) 4.0 V
(c) 5.5 V
Sol: (a) Using Einstein’s photoelectric equation Ek = hν - eVs,
Vs =
(d) 9.5 V
=
= 1.5 V
Q 43: The work function of metal is 3 eV, and then threshold wavelength will be:
(a) 4000 Å
(b) 4133 Å
(c) 4500 Å
Sol: (b) φ = hν0 = hc/λ0 = 3 eV,
(d) 5000 Å
λ0 =
= 3 X 1.6 X 10-19 J
=
m,
4.133 X 10-7 m = 4133 Å
Q 44: A radio station is transmitting the waves of wavelength 300 m. Radiation capacity of the transmitter is 10 kW. Find the no of photons which are
emitted per unit time:
(a) 1.5 × 1029
Sol: (b) p =
(b) 1.5 × 1031
10 kW = 104 J/s,
(c) 1.5 × 1033
n=
Q 45: Photocells are usedfor the:
(b) Reproduction of the pictures from the cinemafilm
=
=
(d) 1.5 × 1035
= 1.5 X 1031
(a) Reproduction of the sound from the cinemafilm
(c) Automatic switching of street lights
(d) Both (a) and (c)
Sol: (d)
Q 46: Light from a hydrogendischargetube is madeincident on the cathodeof a photoelectric cell. The work function of the cathodesurfaceis 3.1 eV. In
order to reducethe photoelectric current to zero value, the minimum potential applied to anodewith respect to cathodeshould be:
(a) – 3.1 V
Sol: (a) Ek = eVs,
(b) – 10.5 V
(c) + 10.5 V
Vs =
J = 3.1 V
(d) – 16.7 V
The stopping potential should be negative so that
it can stop the electrons to reach the
anode.
Q 47: The frequency and work function of an incident photon are v and
. If v0 is the threshold frequency, then necessary condition for the emission of
0
photoelectron is:
(a) v < v0
(c) v ≥ v0
(b) v = v0/2
(d) None of these
Sol: (c)
Q 48: Ultra violet radiation of 6.2 eV falls on an aluminium surface(work function = 4.2 eV). Then kinetic energy in joule of the fastest electron
emitted is approximately:
(a) 3.2 × 10-15
(b) 3.2 × 10-17
Sol: (c) E k = hν - φ, = (6.2 – 4.2) eV,
(c) 3.2 × 10-19
(d) 3.2 × 10-21
= 2 X 1.6 X 10-19 J, = 3.2 X 10-19 J
Q 49: When a monochromatic point sourceof light is at a distanceof 2.0 m from a photocell, the cut – off voltage and the saturation current are
respectively, V 0 = 0.6 V and I s = 18 mA. If the samesourceis placed 0.8 m away from the photocell, then:
(a) Stopping potential V 0 =0.2 V and
saturation current I s =2.0 mA
(b) V 0 =0.6 V and I s =18.0 mA
(c) V 0 =0.6 V and I s =2.0 mA
(d) V 0 =0.2 V and I s =18.0 mA
Sol: (c) When the light sourceis movedaway from the photocell the intensity of light incident on the photocell
number of emitted photoelectrons and hencethe photoelectric current.
decreaseswhich will reducethe
However the frequency (or wavelength) of incident radiations remains
unchangedand so the stopping potential still remains 0.6 V.
Q 50: Threshold wavelength for photoelectric emission from a metal surfaceis 5200 Å. Photoelectrons will be emitted when this surfaceis illuminated
with monochromatic radiation from:
(a) 1 W IR lamp
(b) 50 W UV lamp
(c) 50 W IR lamp
(d) 10 W IR lamp
Sol: (b) For photoelectric emission, the wavelength of incident radiations must be smaller than the threshold
wavelength. Henceonly ultraviolet
lamp is capable of ejecting photoelectrons.
Answers
1. a
2. c
3. b
4. b
5. b
6. b
7. c
8. d
9. b
10. d
11. c
12. c
13. b
14. c
15. d
16. b
17. c
18. a
19. d
20. b
21. c
22. a
23. b
24. b
25. b
26. b
27. a
28. b
29. c
30. b
31. a
32. c
33. a
34. b
35. b
36. d
37. b
38. b
39. d
40. c
41. b
42. a
43. b
44. b
45. d
46. a
47. c
48. c
49. c
50. b
51. d
52. b
53. a
54. a
55. a
56. d
57. a
58. b
59. d
60. a
61. a
62. a
63. a
64. c
65. a
66. d
67. a
68. a
69. d
70. c
71. c
72. b
73. a
74. d
75. a
76. b
77. b
78. b
79. d
80. a