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All questions of Photoelectric Effect for EmSAT Achieve Exam

The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV. What is the stopping potential?​
  • a)
    30 V
  • b)
    5 V
  • c)
    10 V
  • d)
    15 V
Correct answer is option 'C'. Can you explain this answer?

Krishna Iyer answered
Stopping potential (Vo​) is given by
Vo​=W/q​ where W is the work function and q are the charge of an electron.
Given W=20eV−10eV=10eV. Also, q=e
Hence, Vo​=(10eV)/e​=10V

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material .If the frequency is halved and the intensity is doubled, the photoelectric current becomes​
  • a)
    Doubled
  • b)
    Quadrupled
  • c)
    Halved
  • d)
    zero
Correct answer is option 'D'. Can you explain this answer?

Ayush Joshi answered
If the frequency is halved and intensity is doubled, the frequency of incident light will become 15/2 = 0.75 times the threshold frequency. So, as ν<νo Hence, photoelectric current will be zero. 

Photons of energy 6 eV are incident on a potassium surface of a work function 2.1 eV. What is the stopping potential?​
  • a)
    -3.9V
  • b)
    -8.1V
  • c)
    -5V
  • d)
    -1.9V
Correct answer is 'A'. Can you explain this answer?

From photo-electric equation,   eV0​= E−φ
 eV0​=(6−2.1)eV
V0​= 3.9 V
stopping potential is a negative potential to stop e- at saturated current .
 

A photo-sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase the:
  • a)
     voltage applied to the light source
  • b)
     intensity of light
  • c)
     wavelength of light
  • d)
     frequency of light
Correct answer is option 'D'. Can you explain this answer?

Rohan Singh answered
The emission of photoelectron takes place only, when the frequency of the incident light is above a certain critical value, characteristic of that metal. The critical value of frequency is known as the threshold frequency for the metal of the emitting electrode.

Suppose that when light of certain frequency is incident over a metal surface, the photo- electrons are emitted. To take photoelectric current zero, a particular value of stopping potential will be needed. If we go on reducing the frequency of incident light, the value of stopping potential will also go on decreasing. At certain value of frequency v0, the photoelectric current will become zero, even when no retarding potential is applied. This frequency vq corresponds to the threshold for the metal surface. The emission of photoelectrons does not take place, till frequency of incident light is below this value.

Can you explain the answer of this question below:

Light from a bulb is falling on a wooden table but no photo electrons are emitted as

  • A:

    Work function of wood is less

  • B:

    Work function of wood is more

  • C:

    It depends on the frequency

  • D:

    It is independent of work function

The answer is b.

Rishika Patel answered
Explanation:

When light falls on a metal surface, electrons may be emitted from the metal surface. This phenomenon is known as the photoelectric effect. The electrons emitted from the metal surface are called photoelectrons.

The photoelectric effect can be explained by considering that light is made up of photons. Each photon has a certain amount of energy, given by its frequency. When a photon strikes a metal surface, its energy can be transferred to an electron in the metal. If the energy of the photon is greater than the work function of the metal, the electron can be emitted from the metal surface.

In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it. Therefore, no photoelectrons are emitted. This is because the energy of the photons is not enough to overcome the work function of the wood.

Key Points:

- The photoelectric effect is the emission of electrons from a metal surface when light falls on it.
- The energy of a photon is given by its frequency.
- If the energy of a photon is greater than the work function of the metal, electrons can be emitted from the metal surface.
- In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it, so no photoelectrons are emitted.

The electron in a hydrogen atom makes a transition n1 ¾→ n2, where n1 & n2 are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of n1 & n2 are :
  • a)
     n1 = 4, n2 = 2
  • b)
    n1 = 8, n2 = 2 
  • c)
    n1 = 8, n2 = 1
  • d)
    n1 = 6, n2 = 3
Correct answer is option 'A,D'. Can you explain this answer?

Om Desai answered
For an electron revolving in nth orbit around the nucleus of hydrogen atom, Fcentrifugal​=Fcoulomb​
Thus,  mvn2/ rn​​​=​e2​/4πϵo​rn2
Also, mvn​rn​=nh/2π​
On solving these two we get: vn​= e21/2ϵo​hn ​and rn​=( ϵo​h2/πme2 ​) n2 
Time period, T=2πrn/vn​​​⟹T∝n3
Thus, 8T/T​=(​n1/n2)3⟹n1​=2n2
Thus, n1​=4,n2​=2  and   n1​=6,n2​=3 

 In the experiment on photoelectric effect using light having frequency greater than the threshold frequency, the photocurrent will certainly increase when
  • a)
    Anode voltage is incrased
  • b)
    Area of cathode surface is increased
  • c)
    Intensity of incident light is increased
  • d)
    Distance between anode and cathode is increased.
Correct answer is option 'B,C'. Can you explain this answer?

Riya Banerjee answered
If light of certain frequency, greater than the threshold frequency, is incident of the metal surface the photoelectric current obtained is directly proportional to the number of photo-electrons emitted. On increasing the frequency of incident light (or) by increasing the anode potential the speed of photo-electrons emitted changes but they have no effect on the number of electrons emitted. But by increasing the intensity of the incident light more photo-electrons will be emitted and thereby increasing the photoelectric current. 
Also, if we increase the area on which the light is incident we can get more electrons emitted which increases the photoelectric current.
 

When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut off voltage and the saturation current are respectively 0.6 volt and 18.0 mA. If the same source is placed 0.6 m away from the photoelectric cell, then find
Q. 
the saturation current
    Correct answer is '2.0'. Can you explain this answer?

    Neha Sharma answered
    Intensity of photons ↑, no. of electron ↑
    The cut off voltage does not depend on distance between cell and source.
     But intensity is inversely proportional to square of it's distance. 
    ∴ at d=0.6m,
    Cut off voltage =0.6V
    but 
    Is​=18×(0.2/0.6​)2
    =2mA
    So, the answers are option 2.0 mA
     

    In a photo-emissive cell, with exciting wavelength l, the maximum kinetic energy of electron is K. If the exciting wavelength is changed to the kinetic energy of the fastest emitted electron will be :
    • a)
      3K/4
    • b)
      4K/3 
    • c)
      Less than 4K/3
    • d)
      Greater than 4K/3
    Correct answer is option 'D'. Can you explain this answer?

    Lavanya Menon answered
    From E=W0​+(1/2)​mvmax2​⇒vmax​= √[(2E/m)​−(2W0/m)]​​​
    where E= hc​/λ
    If wavelength of incident light charges from λ to 3λ/4​ (Decreases)
    Let energy of incident light charges from E and speed of fastest electron changes from v to v′ then
    v′=√[(2E′/m)​−(2W0/m)​​​]
    As E∝1/λ​⇒E′(4/3)​E
    Hence v′=√[(2(4/3​)E/m)​−(2W0/m)​​]​
    ⇒v′=(4/3)1/2 [(2E/m)​− (2W0/m(4/3)1/2​)]​​
    So, v′>(4/3)1/2v

    An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, then
    • a)
      E = 6.8 eV, l = 6.6 × 10–10 m
    • b)
      E = 3.4 eV, l = 6.6 × 10–10 m
    • c)
      E = 3.4 eV, l = 6.6 × 10–11 m
    • d)
      E = 6.8 eV, l = 6.6 × 10–11 m
    Correct answer is option 'B'. Can you explain this answer?

    Hansa Sharma answered
    The potential energy of an electron is twice the kinetic energy(with negative sign) of an electron.
    PE=2E
    The total energy is: TE=PE+KE
                                     −3.4=−2×3.4+KE
                                     KE=3.4eV
    Let p be the momentum of an electron and m be the mass of an electron.
    E=p2​/2m
    p=√2​mE
     
    Now, the De-Broglie wavelength associated with an electron is:
    λ=h/p​
    λ=h/√2​mE​
    λ=6.6×1034​/√2​×9.1×10−31×(−3.4)×1.6×10−19
    λ=6.6×1034​/9.95×10−25
    0λ=0.66×10−9
    λ=6.6×10−10m

    663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109 incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________.
    • a)
      5.76 x 1011
    • b)
      5.76 x 10
      9
    • c)
      5.76 x 1010
    • d)
      5.76 x 1012
    Correct answer is option 'A'. Can you explain this answer?

    Suresh Iyer answered
    N / Δt = no. of photon incident per second.
    ∴ 663x10-3=(N/Δt)x(hc/λ)
    ∴N/ Δt=663x10-3/(hc/λ)= 663x10-3/(1242nmeV/540)
    (n/Δt)/(N/Δt)=1/(5x109)
    n/Δt=[1/(5x109)]x (N/Δt)
           =[1/(5x109)]x ((663x10-3x540)/1242nmeV)
    I=ne/Δt=5.76x1011A
     

    A particular hydrogen like atom has its ground state binding "energy 122.4 eV. Its is in ground state. Then :
    • a)
      Its atomic number is 3
    • b)
       An electron of 90eV can excite it.
    • c)
      An electron of kinetic energy nearly 91.8 eV can be brought to almost rest by this atom
    • d)
      An electron of kinetic energy 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atom.
    Correct answer is option 'A,C,D'. Can you explain this answer?

    Nikita Singh answered
    En​= −z213.6eV/ n2
    for ground state, n=1
    −122.4=−Z213.6eV
    When a proton interacts that 
    EP​=hV=E1​−E0​=Z213.6eV[(1/12​)−(1/22)​]
    =9×13.6eV [3/4]
    =91.8eV
    So the Atomic number is 3 , a photon  of 91.8eV and an electron of 8.2eV are emitted when 100eV electron interacts e− interact with this atom.

    Two electrons are moving with the same speed v. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field, then after sometime if the de-Broglie wavelengths of the two are l1 and l2 then :
    • a)
      l1 = l2
    • b)
      l1 > l
    • c)
      l1 < l2
    • d)
       l1 > l2 or l1 < l2
    Correct answer is option 'D'. Can you explain this answer?

    Ankita Datta answered
    We know that the de-Broglie wavelength of an electron is given by λ = h/p, where h is the Planck's constant and p is the momentum of the electron.

    In the region of uniform electric field, the electron experiences a force in the direction of the electric field and its momentum changes. However, the magnitude of the momentum remains the same as the speed of the electron remains constant. Therefore, the de-Broglie wavelength of the electron does not change and remains l1.

    In the region of uniform magnetic field, the electron experiences a force perpendicular to its velocity due to the Lorentz force. This force causes the electron to move in a circular path with a radius given by r = mv/qB, where m is the mass of the electron, v is its velocity, q is its charge and B is the magnetic field strength. As the electron moves in a circular path, its momentum changes direction continuously. Therefore, the de-Broglie wavelength of the electron changes and is given by λ2 = h/(mv/qB) = hqB/mv.

    Since the two electrons have the same speed, their momenta are the same. Therefore, we can write:

    λ2 = hqB/mv = (h/mv) * (qB)

    Comparing this with the expression for λ1, we can see that:

    λ2/λ1 = (h/mv) * (qB)/(h/mv) = qB/v

    Therefore, the ratio of the de-Broglie wavelengths of the two electrons depends on the ratio of the magnetic field strength to the speed of the electron. Hence, option (c) is the correct answer.

    Statement-1: In the process of photo electric emission, all the emitted photoelectrons have same K.E.
    Statement-2: According to einstein's photo electric equation KEmax = hn – f.
    • a)
      Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
    • b)
      Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
    • c)
      Statement-1 is true, statement-2 is false.
    • d)
      Statement-1 is false, statement-2 is true.
    Correct answer is option 'D'. Can you explain this answer?

    T.ttttt answered
    When light of frequency f is incident on a metal surface that has a work function W, the maximum kinetic energy of the emitted electrons is given by: 
    KEmax​=hf−ϕ
    Note that this is the maximum possible kinetic energy because ϕ is the minimum energy necessary to liberate an electron.Increasing the frequency of the incident beam, keeping the number of incident photons fixed increases the maximum kinetic energy of the photoelectrons emitted.So, Statement-1 is False, Statement-2 is False.

    In photoelectric effect, stopping potential depends on
    • a)
      Frequency of the incident light
    • b)
      Intensity of the incident light by varies source distance
    • c)
      Emitter's properties
    • d)
      Frequency and intensity of the incident light
    Correct answer is option 'A,C'. Can you explain this answer?

    Rutuja Ahuja answered
    Explanation:
    The photoelectric effect is a phenomenon where electrons are emitted from the surface of a metal when light of a certain frequency or higher is incident on it. The stopping potential is the minimum potential required to stop the electrons emitted from the metal surface. The stopping potential depends on the frequency of the incident light and the emitter's properties.

    Frequency of the incident light:
    The stopping potential is directly proportional to the frequency of the incident light. This is because the energy of a photon is directly proportional to its frequency. The work function of the metal is the minimum energy required to remove an electron from its surface. If the energy of the incident photon is greater than the work function, then an electron will be emitted. The remaining energy of the photon will contribute to the kinetic energy of the emitted electron. Therefore, increasing the frequency of the incident light will increase the kinetic energy of the emitted electrons, and hence, the stopping potential required to stop them.

    Emitter's properties:
    The stopping potential also depends on the emitter's properties, such as its work function and the number of free electrons available for emission. The work function is the minimum energy required to remove an electron from the surface of the metal. Metals with a higher work function require more energy to remove an electron, and hence, a higher stopping potential to stop the emitted electrons. The number of free electrons available for emission also affects the stopping potential. Metals with a higher number of free electrons will require a higher stopping potential to stop the emitted electrons.

    Intensity of the incident light:
    The intensity of the incident light does not affect the stopping potential. This is because the number of emitted electrons is directly proportional to the intensity of the incident light. However, the kinetic energy of the emitted electrons will increase with increasing intensity, as more energy is transferred to each electron.

    Frequency and intensity of the incident light:
    The stopping potential depends only on the frequency of the incident light and the emitter's properties, not on the intensity of the incident light. Therefore, the correct answer is option A,C.

    Light coming from a discharge tube filled with hydrogen falls on the cathode of the photoelectric cell. The work function of the surface of cathode is 4eV. Which one of the following values of the anode voltage (in Volts) with respect to the cathode will likely to make the photo current zero.
    • a)
      –4
    • b)
      –6
    • c)
      –8
    • d)
      –10
    Correct answer is option 'D'. Can you explain this answer?

    Pragati Dey answered
    Energy of photon from H atom is gievn by the difference between the two energy level in which electron transit.for minimum wavelength, energy of the photon will be maximum.

    maximum energy will be

    according to the einstein equation,Maximum kinetic energy if the emitted electron = energy of the incident radiation − work function
    KE = 13.6 − 4 = 9.6 eV
    hence potenial of the node must be greater than or equal to 9.6 volt with negatice polarity.
    Hnece option (d) is correct.

     A beam of ultraviolet light of all wavelengths passes through hydrogen gas at room temperature, in the x-direction. Assume that all photons emitted due to electron transition inside the gas emerge in the y-direction. Let A and B denote the lights emerging from the gas in the x and y directions respectively.
    • a)
      Some of the incident wavelengths will be absent in A
    • b)
      Only those wavelengths will be present in B which are absent in A
    • c)
      B will contain some visible light.
    • d)
      B will contain some infrared light.
    Correct answer is option 'A,C,D'. Can you explain this answer?

    Harsh Desai answered
    Given, incident wavelengths travel in x-direction which are in the U.V region, which means they can excite an electron from n=1 to higher levels. Hence, some of the incident wavelengths will be absent in A.
    Photons emerging due to transitions of electrons to their ground state contains visible region wavelengths if transition is done into n=2. If transition is done into n>2, photons contain I.R wavelengths.
    Therefore, A,C,D is the correct choice.
     

    Statement-1: Work function of aluminum is 4.2 eV. If two photons each of energy 2.5 eV strikes on a piece of aluminum, the photo electric emission does not occur.
    Statement-2: In photo electric effect a single photon interacts with a single electron and electron is emitted only if energy of each incident photon is greater then the work function.
    • a)
      Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
    • b)
       Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
    • c)
      Statement-1 is true, statement-2 is false.
    • d)
      Statement-1 is false, statement-2 is true.
    Correct answer is option 'A'. Can you explain this answer?

    Statement-1: Work function of aluminum is 4.2 eV. If two photons each of energy 2.5 eV strikes on a piece of aluminum, the photoelectric emission does not occur.
    Statement-2: In the photoelectric effect, a single photon interacts with a single electron, and the electron is emitted only if the energy of each incident photon is greater than the work function.

    To analyze the given statements, let's break them down and discuss each one separately.

    Statement-1: Work function of aluminum is 4.2 eV. If two photons each of energy 2.5 eV strikes on a piece of aluminum, the photoelectric emission does not occur.

    The work function of a material is the minimum amount of energy required to remove an electron from the material's surface. In this case, the work function of aluminum is stated as 4.2 eV. When two photons, each with an energy of 2.5 eV, strike the aluminum surface, the total energy of the photons is 5 eV (2.5 eV + 2.5 eV).

    Statement-2: In the photoelectric effect, a single photon interacts with a single electron, and the electron is emitted only if the energy of each incident photon is greater than the work function.

    The photoelectric effect is the phenomenon where electrons are emitted from a material's surface when it is exposed to light. According to the statement, a single photon interacts with a single electron, and the electron is emitted only if the energy of the incident photon is greater than the work function of the material.

    Explanation:
    Both statements are true, and statement-2 provides a correct explanation for statement-1. Here's why:

    When the two photons with energies of 2.5 eV each strike the aluminum surface, their total energy is 5 eV (2.5 eV + 2.5 eV). In the photoelectric effect, the energy of each incident photon should be greater than the work function for electron emission to occur.

    In this case, the energy of each incident photon (2.5 eV) is less than the work function of aluminum (4.2 eV). Therefore, even though the total energy of the two photons is sufficient (5 eV), the energy of each individual photon is not enough to overcome the work function barrier. As a result, the photoelectric emission does not occur.

    Thus, statement-2 correctly explains why the photoelectric emission does not occur in this scenario, supporting statement-1.

    In conclusion, both statements are true, and statement-2 provides a correct explanation for statement-1.

    If radiation of allow wavelengths from ultraviolet to infrared is passed through hydrogen agas at room temperature, absorption lines will be observed in the :
    • a)
      Lyman series
    • b)
       Balmer series
    • c)
      Both (A) and (B)
    • d)
      Neither (A) nor (B)
    Correct answer is option 'A'. Can you explain this answer?

    Shilpa Saha answered
    At room temperature, nearly all the atoms in hydrogen gas will be in the ground state. When light passes through the gas, photons are absorbed, causing electrons to make transitions to higher states and creating absorption lines. These lines correspond to the Lyman series since that is the series of transitions involving the ground state or n = 1 level. Since there are virtually no atoms in higher energy states, photons corresponding to transitions from n > 2 to higher states will not be absorbed.

    Statement-1: An electron and a proton are accelerated through the same potential difference. The deBroglie wavelength associated with the electron is longer.
    Statement-2: De-Broglie wavelength associated with a moving particle is l =  where, p is the linear momentum and both have same K.E.
    • a)
      Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
    • b)
      Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
    • c)
      Statement-1 is true, statement-2 is false.
    • d)
      Statement-1 is false, statement-2 is true.
    Correct answer is option 'A'. Can you explain this answer?

    Shubham Jain answered
    de Broglie Wavelength for Electrons
    In the case of electrons going in circles around the nuclei in atoms, the de Broglie waves exist as a closed-loop, such that they can exist only as standing waves, and fit evenly around the loop. Because of this requirement, the electrons in atoms circle the nucleus in particular configurations, or states, which are called stationary orbits.
    de Broglie Wavelength Formula and Derivation
    de Broglie reasoned that matter also can show wave-particle duality, just like light, since light can behave both as a wave (it can be diffracted and it has a wavelength) and as a particle (it contains packets of energy hν). And also reasoned that matter would follow the same equation for wavelength as light namely,
    λ = h / p
    Where p is the linear momentum, as shown by Einstein.
    Derivation
    de Broglie derived the above relationship as follows:
    1) E = hν for a photon and λν = c for an electromagnetic wave.
    2) E = mc2, means λ = h/mc, which is equivalent to λ = h/p.
    Note: m is the relativistic mass, and not the rest mass; since the rest mass of a photon is zero.
    Now, if a particle is moving with a velocity v, the momentum p = mv and hence λ = h / mv
    Therefore, the de Broglie wavelength formula is expressed as;
    λ = h / mv

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