Q2: The graph which shows the variation of (1 / λ^{2}) and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle):
(a)
(b)
(c)
(d) [2024]
Ans: (d)
deBroglie wavelength
Squaring both sides,
Graph passes through origin with constant slope.
Q1: The work functions of Caesium (Cs), Potassium (K), and Sodium (Na) are 2.14 eV, 2.30 eV, and 2.75 eV respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?
A: Cs only
B: Both Na and K
C: K only
D: Na only
Ans: A
Solution: Energy of incident radiation = 2.80 eV
The work function of Cs → 2.14 eV
The work function of K → 2.30 eV
The work function of Na → 2.75 eV
Since the work functions of potassium and sodium are more than the energy of incident radiation hence photons may be emitted from cesium.
Q2: The minimum wavelength of Xrays produced by an electron accelerated through a potential difference of V volts is proportional to :
(a) 1/V
(b) 1/√V
(c) V^{2}
(d) √V
Ans: (a)
Minimum wavelength of XRays is
Q1: The light rays having photons of energy 4.2 eV are falling on a metal surface having a work function of 2.2 eV. The stopping potential of the surface is
(a) 6.4 V
(b) 2 eV
(c) 2 V
(d) 1.1 V
Ans: (c)
Q2: The threshold frequency of a photoelectric metal is v_{0}. If light of frequency 4v_{0} is incident on this metal, then the maximum kinetic energy of emitted electrons will be :
(a) 4 hv_{0}
(b) hv_{0}
(c) 2 hv_{0}
(d) 3 hv_{0}
Ans: (d)
According to Einstein's photoelectric equation
Q3: When two monochromatic lights of frequency, ν and ν/2 are incident on a photoelectric metal, their stopping potential becomes Vs/2 and Vs respectively. The threshold frequency for this metal is
A: 3ν
B: 2/3ν
C: 3/2ν
D: 2ν
Ans: (c)
Q1: The number of photons per second on average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3 × 10^{–3} watts will be : (h = 6.6 × 10^{–34 }Js)
A: 10^{16}
B: 10^{15}
C: 10^{18}
D: 10^{17}
Ans: A
Solution:
Q2: An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' is the mass of photoelectron emitted from the surface that has deBroglie wavelength λd, then:
A:
B:
C:
D:
Ans: A
Solution:
Q1: Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and the intensity is doubled?
A: onefourth
B: Zero
C: doubled
D: four times
Ans: B
Solution:
f_{0} < f_{1} = 1.5 f_{0}
∴ f_{2} = 0.75 f_{0}
for given condition
f_{incident }< f_{threshold}
so no photo electron emission
i = 0
Q2: The deBroglie wavelength of an electron moving with kinetic energy of 144 eV is nearly
(a) 102 × 10^{−3} nm
(b) 102 × 10^{−4} nm
(c) 102 × 10^{−5} nm
(d) 102 × 10^{−2} nm
Ans: (d)
Kinetic energy of electron, K = 144 eV
⇒ eV = 144 eV
⇒ V = 144 V
∴ deBroglie wavelength
Q3: The wave nature of electrons was experimentally verified by
(a) deBroglie
(b) Hertz
(c) Einstein
(d) Davisson and Germer
Ans: (a)
The wave nature of electrons was experimentally verified by deBroglie.
Q1: An electron is accelerated through a potential difference of 10,000 V. Its deBroglie wavelength is, (nearly) : (m_{e} = 9 × 10^{–31} kg)
A: 12.2 × 10^{–13} m
B: 12.2 × 10^{–12} m
C: 12.2 × 10^{–14} m
D: 12.2 nm
Ans: B
For an electron accelerated through a potential V
Q2: The work function of a photosensitive material is 4.0 eV.
This longest wavelength of light that can cause photon emission from the substance is (approximately)
(a) 3100 nm
(b) 966 nm
(c) 31 nm
(d) 310 nm
Ans: (d)
The work function of material is given by
where, h = Planck’s constant = 6.63 × 10^{−34} Js
c = speed of length = 3 × 10^{8} ms^{−1}
and λ = wavelength of light
Here, φ = 4 eV = 4 × 1.6 × 10 ^{−19} J
Substituting the given values in Eq. (i), we get
Q3: An electron is accelerated through a potential difference of 10,000 V.
Its deBroglie wavelength is, (nearly) : (m e = 9 × 10^{−31} kg)
(a) 12.2 × 10^{−12} m
(b) 12.2 × 10^{−14}m
(c) 12.2 nm
(d) 12.2 × 10^{−13}m
Ans: (a)
Given, potential difference, V = 10000 V If electron is accelerated through a potential of V volt, then the wavelength associated with it is given by
where, h = Planck’s constant = 6.63 × 10^{−34} Js,
e = electronic charge = 1.6 × 10^{−19 }C a
nd m_{e} = mass of electron = 9 × 10^{−31} kg Substituting these values in Eq. (i),
we get
Q1: An electron of mass m with an initial velocityenters an electric field E_{0} = constant > 0) at t = 0. If λ_{0} is its deBroglie wavelength initially, then its deBroglie wavelength at time t is:
A:
B:
C: λ_{0}t
D: λ_{0}
Ans: A
Q2: When the light of frequency 2v_{0} (where v_{0} is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v_{1}. When the frequency of the incident radiation is increased to 5 v_{0}, the maximum velocity of electrons emitted from the same plate is v_{2}. The ratio of v_{1} to v_{2} is :
A: 1 : 2
B: 1 : 4
C: 4 : 1
D: 2 : 1
Ans: A
Solution:
Q3: When the light of frequency 2ν_{0} (where, ν_{0} is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v_{1} . When the frequency of the incident radiation is increased to 5ν_{0} , the maximum velocity of electrons emitted from the same plate is v_{2} . The ratio of v_{1} to v_{2} is
(a) 4 : 1
(b) 1 : 4
(c) 1 : 2
(d) 2 : 1
Ans: (c)
According to the Einstein’s photoelectric equation,
where, K_{max} is the maximum kinetic energy of photoelectrons having maximum velocity v_{max}. When incident frequency of light, v = 2v_{0} Substituting the value of ν in Eq. (i), we get
If incident frequency of radiation, ν = 5ν_{0} Substituting the value of ν in Eq. (i), we get
On dividing Eq. (ii) by Eq (iii), we get
Q1: The photoelectric threshold wavelength of silver is 3250 × 10^{–10}m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10–10 m is:
(Given h = 4.14 × 10^{–15} eVs and c = 3 × 10^{8} ms^{–1})
A: ≈ 0.6 × 10^{6} ms^{–1}
B: ≈ 61 × 10^{3} ms^{–1}
C: ≈ 0.3 × 10^{6} ms^{–1}
D: ≈ 6 × 10^{5} ms^{–1}
Ans: A
Solution:
Q2: The deBroglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is
(a)
(b)
(c)
(d)
Ans: (b)
Thinking Process deBroglie wavelength associated with a moving particle can be given as
At thermal equilibrium, temperature of neutron and heavy water will be same. This common temperature is given as, T. Also, we know that, kinetic energy of a particle
where, p = momentum of the particle m = mass of the particle Kinetic energy of the neutron is
K.E. = 3/2 kT
∴ deBroglie wavelength of the neutron
Q1: When a metallic surface is illuminated with radiation of wavelength λ the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. The threshold wavelength for the metallic surface is :
A: 3λ
B: 4λ
C: 5λ
D: 5/2λ
Ans: A
When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V.
The photoelectric equation can be written as,
Now, when the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. So, the photoelectric equation can be written as,
From equations (i) and (ii), we get
When a metallic surface is illuminated with radiation of wavelength λ stopping potential is V.
The photoelectric equation can be written as,
Now, when the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. So, the photoelectric equation can be written as,
From equations (i) and (ii), we get
Q2: An electron of mass m and a photon have the same energy E. The ratio of deBroglie wavelengths associated with them is:
A:
B:
C:
D: c(2mE)^{1/2}
Ans: B
Given that an electron has a mass of m.
DeBroglie wavelength for an electron will be given as,
where,
h is the Planck's constant, and
p is the linear momentum of the electron
The kinetic energy of the electron is given by,
From equation (i) and (ii), we have
The energy of a photon can be given as,
Hence, λ_{P} is the deBroglie wavelength of the photon.
Now, dividing equation (iii) by (iv), we get
Q3: Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is
(a) + 3 V
(b) + 4 V
(c) – 1 V
(d) – 3 V
Ans: (d)
Key Idea Use Einstein’s photoelectric equation.
We know that, E = (KE)_{max }+ Work function (φ)
where, φ = hν_{0} E = hν
Q4: Electrons of mass m with deBroglie wavelength λ fall on the target in an Xray tube. The cutoff wavelength (λ 0 ) of the emitted Xray is
Ans: (a)
Cutoff wavelength occurs when incoming electron looses its complete energy in collision. This energy appears in the form of Xrays.
Given, mass of electrons = m
deBroglie wavelength = λ
So, kinetic energy of electron = p^{2}/2m
Now, maximum energy of photon can be given by
Q1: A certain metallic surface is illuminated with monochromatic light of wavelength, λ. The stopping potential for photoelectric current for this light is 3V_{0}. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V_{0}. The threshold wavelength for this surface for the photoelectric effect is :
A: λ/6
B: 6λ
C: 4λ
D: λ/4
Ans: C
We have,
where W is the work function and (3V_{0}) is the stopping potential when monochromatic light of wavelength λ is used.
where V_{0} is the stopping potential when monochromatic light of wavelength 2λ is used.
Subtracting equation (2) from equation (1)
We get,
∴
Substituting in equation (2) we get,
∴
The threshold wavelength is therefore 4λ.
Q2: Which of the following figures represents the variation of the particle momentum and the associated deBroglie wavelength?
A:
B:
C:
D:
Ans: C
The deBroglie wavelength is given by
This equation is in the form of yx = c, which is the equation of a rectangular hyperbola.
The deBroglie wavelength is given by
This equation is in the form of yx = c, which is the equation of a rectangular hyperbola.
Q1: When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increases from 0.5 eV to 0.8 eV. The work function of the metal is :
A: 1.3 eV
B: 1.5 eV
C: 0.65 eV
D: 1.0 eV
Ans: D
Solution:
The original energy of the photon is E_{0}
From equation (i) and (ii)
Q2: If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de−Broglie wavelength of the particle is :
A: 60
B: 50
C: 25
D: 75
Ans: D
Solution:
Q3: Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The deBroglie wavelength of the emitted electron is
(a) < 2.8 × 10^{− 10} m
(b) < 2.8 × 10^{− 9} m
(c) ≥ 2.8 × 10^{− 9} m
(d) ≤ 2.8 × 10^{− 12} m
Ans: (c)
As, energy of photon, E = hν
According to Einstein’s photoelectric emission, we have
KE_{max} = E − W = 2.48 − 2.28 = 0.2 eV
For deBroglie wavelength of the emitted electron,
Thus, minimum wavelength of the emitted electron is
105 videos425 docs114 tests

1. What is the photoelectric effect? 
2. How does the photoelectric effect support the waveparticle duality of light? 
3. What is the threshold frequency in the context of the photoelectric effect? 
4. How does the kinetic energy of emitted electrons in the photoelectric effect depend on the frequency of light? 
5. What is the significance of the work function in the photoelectric effect? 

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