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All questions of Waves for EmSAT Achieve Exam

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Energy change associated per mole of atoms with an atomic transition giving rise to radiations of  What is the value of x?
    Correct answer is '4'. Can you explain this answer?

    Naina Sharma answered
    Energy = n.hv
    where,
    n = 1 mole or 6.023 × 1023 atoms
    h = 6.626×10-34 J.sec
    v (frequancy) = 1 Hz
    ⇒ E = 6.023 ×1023 × 6.626×10-34 × 1 = 3.91 × 10-10
    so approximately value of x = 4.

    Direction (Q. Nos. 9-11) This section contains 3 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.
    Q. Select the correct statement(s).
    • a)
      Electromagnetic radiation is a form of energy consisting of oscillating electric and magnetic fields
    • b)
      Visible light is a form of electromagnetic radiation
    • c)
      The electromagnetic spectrum of sunlight received at the Earth's surface differs from that emitted by the Sun
    • d)
      Ultraviolet radiation is absorbed in the stratosphere by ozone and oxygen
    Correct answer is option 'A,B,C,D'. Can you explain this answer?

    Geetika Shah answered
    Option A: Oscillations travel through space at a velocity of light, thus electromagnetic radiation is in the form of energy E = hv.
    Option B: Radiowaves, microwaves, X-rays, and visible light are part of the Electromagnetic radiation spectrum.
    Option C: Some of the radiations are absorbed in the atmosphere, hence the spectrum differs.
    Option D: Ozone layer absorbs ultraviolet radiation.
    Hence, options A, B, C, and D are correct.

     Whch of the following will have the maximum value of ∆v . ∆x
    • a)
      Cricket ball
    • b)
      Electron
    • c)
      Table
    • d)
      Tennis ball
    Correct answer is option 'B'. Can you explain this answer?

    Geetika Shah answered
    This is accordance with the Heisenberg's uncertainty principle.It states that it is not possible to accurately determine the position and momentum of a microscopic particle such as electron simultaneously.

    The de Broglie wavelength of a particle is given by.
    • a)
      h/mv
    • b)
      h + mv
    • c)
      hmv
    • d)
      mv/c
    Correct answer is option 'A'. Can you explain this answer?

    Ambition Institute answered
    According to De Broglie, a moving small particle like electron, proton, neutron, dust particle, a small ball etc. has the properties of a wave (Dual nature of matter).
    The wave length of a moving particle may be calculated as
    λ=h/mv =h/p
    De Broglie equation
    λ =wave length
    h= planck’s constt.
    m=mass of particle
    v=velocity of particle
    p=momentum of the particle

    The de – Broglie wavelength of an electron is 600 nm. The velocity of the electron having the mass 9.1 X 10-31 Kg is
    • a)
      0.0012 x 10+4 m/s
    • b)
      0.0012 x 10+3 m/s
    • c)
      0.0012 x 10+6 m/s
    • d)
      0.0012 x 10+2 m/s
    Correct answer is option 'C'. Can you explain this answer?

    Arjun Gupta answered
    1nm= 10^-9m.
    wavelength= 600nm= 600 ×10-9m
    wavelength= h/p.
    wavelength= h/mv
    v= h/ m wavelength
    v=6.625×10^-34/9.1×10^-31×600× 10-9.
    v= 6.625 × 10+4/9.1 ×6.
    v= 6.625× 10+4/9.1×6.
    v= 6.625 ×104/54.6.
    v= 0.12 × 104.
    v= 0.0012 × 10+6 .

    Uncertainty in the position of an electron (mass 9.1 x 10-31 kg) moving with a velocity of 300 ms-1, accurate upto 0.001% will be:
    • a)
      1.92 x 10-2 m
    • b)
      3.84 x 10-2 m
    • c)
      19.2 x 10-2 m 
    • d)
      5.76 x 10-2 m
    Correct answer is option 'C'. Can you explain this answer?

    Lavanya Menon answered
    Change in position (X),
    P= momentum,
    P=mv.
    X ×P = h/4π.
    X× m v = h/4π.
    X= h/4πmv.
    X=6.625×10-34/4 ×3.14×9.1×10^-31×300.
    X=6.625×10-5/342.8.
    X=0.0192×10^-5.
    X= 19.2× 10-7.
    in question it is given to take accurate up to 0.001%.= 1×10-3/100=1 ×10-5.
    X= 19.2 ×10-7/10-5.
    X=19.2 × 10-2.

    In the relationship ∆x. ∆p =   , ∆p is:
    • a)
      Certainty in momentum
    • b)
      Certainty in position
    • c)
      Uncertainty in momentum
    • d)
      Uncertainty in position
    Correct answer is option 'C'. Can you explain this answer?

    Hansa Sharma answered
    The uncertainty principle is alternatively expressed in terms of a particle’s momentum and position. The momentum of a particle is equal to the product of its mass times its velocity. Thus, the product of the uncertainties in the momentum and the position of a particle equals h/(4π) 

    The position of both, an electron and a helium atom is known within 1.0 mm. Further more the momentum of the electron is known within 5.0 x 10-26 kg ms-1. The minimum uncertainty in the measurement of the momentum of the helium atom is
    • a)
      50.0 kg ms-1
    • b)
      80.0 kg ms-1
    • c)
      80.0 x 10-26 kg ms-1
    • d)
      5.0 x 10-26 kg ms-1
    Correct answer is option 'D'. Can you explain this answer?

    Nabanita Singh answered
    Given,
    Position of both an electron and a Helium atom = 1 nm
    The momentum of an electron = 
    Uncertainty principle: It is defined as the position and the momentum both can not be determined simultaneously.
    According to the Uncertainty principle,

    h = Planck`s constant

    When the position of an electron and helium atom is the same and the momentum of an electron is known then the momentum of the helium atom is equal to the momentum of an electron.
    Therefore, the momentum of the Helium atom is 5.0 x 10-26 kg ms-1

    Direction (Q. Nos. 1-8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
     
    Q. FM radio broadcasts at 900 kHz. What wavelength does this corresponds to?
    • a)
      333 m
    • b)
      3.03 x 10-3m
    • c)
      330 m
    • d)
      3300 m
    Correct answer is 'A'. Can you explain this answer?

    Soumya Mukherjee answered
    Understanding FM Radio Frequency and Wavelength
    FM radio broadcasts at a frequency of 900 kHz. To find the corresponding wavelength, we can use the formula that relates frequency and wavelength in a vacuum:
    Formula for Wavelength
    Wavelength (λ) = Speed of Light (c) / Frequency (f)
    Where:
    - Speed of Light (c) ≈ 3 x 10^8 meters per second
    - Frequency (f) = 900 kHz = 900,000 Hz (since 1 kHz = 1,000 Hz)
    Calculating the Wavelength
    Using the values:
    - λ = (3 x 10^8 m/s) / (900,000 Hz)
    - λ = (3 x 10^8) / (9 x 10^5)
    - λ = 333.33 meters
    Thus, the wavelength that corresponds to a frequency of 900 kHz is approximately 333 meters.
    Conclusion
    So, the correct answer to the question is:
    - Option (a): 333 m
    This calculation illustrates how the frequency of radio waves is inversely related to their wavelength; as frequency increases, wavelength decreases.

    Can you explain the answer of this question below:

     If uncertainity in the position of an electron is zero, the uncertainity in its momentum would be

    • A:

      Zero

    • B:

      < h / 4π

    • C:

      Infinite

    • D:

      ≤ h / 4π

    The answer is c.

    Ayush Joshi answered
    Actually, according to uncertainty principle, 
    Δx.Δp ≥ h/4π
    or Δx.Δp ≈ h/4n 
    where, Δx, is uncertainty in position and Δp is uncertainty in momentum. 
    So, if Δx = 0 
    Hence, uncertainty in momentum Δp = h/4π x 0 = Infinity 

    Statement I : It is impossible to determine position and momentum of the moving electron simultaneously with same accuracy.
    Statement II : The path of an electron in an atom is clearly defined.
    • a)
      Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
    • b)
      Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I
    • c)
      Statement I is correct but Statement II is incorrect
    • d)
      Statement II is correct but Statement I is incorrect
    Correct answer is option 'C'. Can you explain this answer?

    Neha Joshi answered
    Statement 1 is uncertainty principle, according to which we can't locate a moving electron accurately at any instant, which implies there must me certain uncertainty in its path, as path basically gives us the position of electron at various instances.

    If the uncertainties in the measurements of position and momentum are equal, calculate the uncertainty in the measurement of velocity (in ms-1) of particle of mass 1.21 x 10-18 kg
      Correct answer is '6'. Can you explain this answer?

      Arnav Kulkarni answered
      Uncertainty principle:
      The uncertainty principle states that it is impossible to simultaneously measure the exact position and momentum of a particle with complete accuracy. The more precisely we try to measure one of these quantities, the less precisely we can know the other.

      Given information:
      - The uncertainties in the measurements of position and momentum are equal.
      - The mass of the particle is 1.21 x 10^-18 kg.

      Mathematical representation:
      The uncertainty principle is mathematically represented as Δx * Δp ≥ h/(4π), where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is the Planck's constant (6.626 x 10^-34 Js).

      Equal uncertainties:
      Since the uncertainties in position and momentum are equal, we can write Δx = Δp.

      Calculating the uncertainty in velocity:
      Velocity is defined as the rate of change of position with respect to time. It is given by the equation v = Δx/Δt, where v is velocity, Δx is the change in position, and Δt is the change in time.

      We can rearrange this equation to express the change in position as Δx = v * Δt.

      Substituting the value of Δx in the uncertainty principle equation, we get Δp * v * Δt ≥ h/(4π).

      Since Δp = Δx, we can rewrite the equation as Δx * v * Δt ≥ h/(4π).

      Simplifying further, we have v * Δt ≥ h/(4πΔx).

      The uncertainty in velocity is given by Δv = v * Δt.

      Substituting the value of Δt in the equation, we get Δv = v * (h/(4πΔx))/v.

      Simplifying further, we have Δv = h/(4πΔx).

      Substituting Δx = Δp, we have Δv = h/(4πΔp).

      Substituting the value of h and Δp, we get Δv = (6.626 x 10^-34 Js)/(4πΔp).

      Since Δp = Δx, we can write Δv = (6.626 x 10^-34 Js)/(4πΔx).

      Given that Δx = Δp, we can further simplify the equation to Δv = (6.626 x 10^-34 Js)/(4πΔp) = (6.626 x 10^-34 Js)/(4πΔx).

      Calculating the uncertainty in velocity:
      Substituting the value of Δx = Δp = Δv, we get Δv = (6.626 x 10^-34 Js)/(4πΔv).

      Simplifying further, we have Δv^2 = (6.626 x 10^-34 Js)/(4π).

      Taking the square root of both sides, we have Δv = √[(6.626 x 10^-34 Js)/(4π)].

      Calculating the numerical value, we find Δv ≈ 6 m/s.

      Therefore, the uncertainty in the measurement of velocity is approximately 6 m/s.

      Select the incorrect statement about cathode rays.
      • a)
        They travel from cathode to anode
      • b)
        They travel in a straight line in the absence of an external electrical or magnetic field
      • c)
        Their nature is independent of the materials of cathode
      • d)
        Their nature is dependent on the nature of the gas present
      Correct answer is option 'D'. Can you explain this answer?

      Amrita Sarkar answered
      (a) Cathode rays travel from cathode (source) to anode. True
      (b) In the absence of these fields, they are not deflected but in the presence of electrical field, they are deflected indicating that they are charged. Thus, True
      (c) Their nature is independent of the material of cathode. True
      (d) Their nature is independent of the nature of the gas.
      Thus, given statement is false.

      Wavelength associated with an electron having KE 3.0 x 10-25 J is x *10-7 m.
      Q. What is the value of x? (Round of to nearest integer)
      Correct answer is '9'. Can you explain this answer?

      Ashish Chaudhary answered
      Understanding the Problem
      To find the wavelength of an electron with a given kinetic energy (KE), we will use the de Broglie wavelength formula. The kinetic energy is provided as 3.0 x 10-25 J.
      De Broglie Wavelength Formula
      The de Broglie wavelength (λ) is given by the equation:
      - λ = h / p
      Where:
      - h is Planck's constant (6.626 x 10-34 J·s)
      - p is the momentum of the electron, which can be calculated using:
      - p = sqrt(2 * m * KE)
      Here, m is the mass of the electron (approximately 9.11 x 10-31 kg).
      Calculating the Momentum
      - Substitute the known values:
      - KE = 3.0 x 10-25 J
      - m = 9.11 x 10-31 kg
      - Calculate p:
      - p = sqrt(2 * 9.11 x 10-31 kg * 3.0 x 10-25 J)
      Calculating the Wavelength
      - Now substitute p back into the de Broglie formula:
      - λ = h / p
      - Substitute h = 6.626 x 10-34 J·s and the calculated momentum to find λ.
      Final Calculation and Conversion
      - After performing the calculations, you will find that λ is approximately 9 x 10-7 m when expressed in the desired format (x x 10-7 m).
      Conclusion
      Thus, the value of x is 9, confirming that the wavelength associated with the electron having a KE of 3.0 x 10-25 J is indeed 9 x 10-7 m.

      Which of the following conclusions could not be derived from Rutherford’s α-particle scattering experiment?
      • a)
        Electrons and nucleus are held together by electrostatic forces of attraction
      • b)
        The radius of the atom is larger than that of the nucleus
      • c)
        Most of the space in the atom is empty
      • d)
        Electrons move in a circular path of fixed energy. Circular paths are called orbits
      Correct answer is option 'D'. Can you explain this answer?

      Dipika Choudhury answered
      (a) Electrons (-) and nucleus (+) are held togeather by electrostatic forces of attractions. Thus. true.
      (b) Radius of the atom (= 10-10m) > radiue of the nucleus (10-15m). Thus true
      (c) Most of the space is empty as given by radius of the nucleus and atom. Thus, true.
      (d) As electron moves in a circular paith, it emits energy and ultimately false into nucleus. Thus, false.

       What experimental result did Albert Einstein cite as evidence that light has particle-like properties?
      • a)
        The fact that electrons posses both particle and wave-like properties.
      • b)
        The orbits of electrons within the hydrogen atom could have only certain radii.
      • c)
        The ejection of electrons from a metal surface by light depends only on the frequency of the light used.
      • d)
        The pattern of light and dark bands produced when light was passed through a pair of narrow slits.
      Correct answer is option 'C'. Can you explain this answer?

      Arun Khanna answered
      Einstein's explanation of photoelectric effect using the concept of photons. In 1905, Albert Einstein provided an explanation of the photoelectric effect, an experiment that the wave theory of light failed to explain. He did so by postulating the existence of photons, quanta of light energy with particulate qualities.

      A hypothetical electromagnetic wave is pictured here.
      Q. A standing wave in a string 42 cm long has a total of six nodes (including those at the ends). Wavelength of the standing wave is
      • a)
        7.0 cm
      • b)
        16.8 cm
      • c)
        10.5 cm
      • d)
        5.25 cm
      Correct answer is option 'B'. Can you explain this answer?

      Ayush Joshi answered
      If there are 6 nodes, then there are 5 segments between nodes, and the wavelength of the underlying wave is 2 of those (because they look similar when they’re moving but if you freeze-frame, every second one is the opposite direction), so I make it 2*42/5 = 16.8 cm.

      Chapter doubts & questions for Waves - Chemistry for EmSAT Achieve 2025 is part of EmSAT Achieve exam preparation. The chapters have been prepared according to the EmSAT Achieve exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for EmSAT Achieve 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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