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Karnataka PUC II Dual Nature of Radiation and Matter — practice questions — page 2

Explore 98 free Dual Nature of Radiation and Matter questions in Karnataka PUC II Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Dual Nature of Radiation and Matter questions and exercises with answers

Showing questions 31–60 of 98.

For accelerated electron with KE = 100 eV, de Broglie wavelength:Read the answer and worked solution →Energy of photon with momentum 3.3 × 10⁻²⁵ kg m/s:Read the answer and worked solution →For photoelectric experiment, plot of KE_max vs frequency ν is:Read the answer and worked solution →In Compton scattering at 90° (cos 90 = 0), wavelength shift:Read the answer and worked solution →Maximum velocity of photoelectrons from a metal of work function 2 eV, illuminated by 200 nm light:Read the answer and worked solution →In a photoelectric experiment, the minimum negative potential applied to the collector plate at which the photocurrent just drops to zero is called theRead the answer and worked solution →The momentum carried by a photon of wavelength $\lambda$ isRead the answer and worked solution →NEET · Physics · Photoelectric — Spot the ErrorRead the answer and worked solution →Knock out the photoelectric mythsRead the answer and worked solution →The minimum energy needed to free an electron from a metal is its:Read the answer and worked solution →Work function is usually measured in:Read the answer and worked solution →Photoelectric emission was discovered in 1887 by:Read the answer and worked solution →Below the threshold frequency, photoelectric emission:Read the answer and worked solution →The maximum photoelectric current at high positive voltage is the:Read the answer and worked solution →The negative voltage that just stops the current is called the:Read the answer and worked solution →Photoelectric emission begins after a time lag that is:Read the answer and worked solution →Who explained the photoelectric effect in 1905?Read the answer and worked solution →In Einstein's picture, radiation travels as:Read the answer and worked solution →The number of photoelectrons emitted per second depends on the:Read the answer and worked solution →The kinetic energy of the photoelectrons is set by the:Read the answer and worked solution →Increasing the intensity of light leaves the stopping potential:Read the answer and worked solution →The work function of a surface depends on the:Read the answer and worked solution →Who proposed that matter also has a wave nature?Read the answer and worked solution →The waves associated with moving particles are called:Read the answer and worked solution →The de Broglie relation contains a wave concept together with a:Read the answer and worked solution →Ultraviolet light causing emission was studied in detail by:Read the answer and worked solution →However intense the light, no emission occurs below the:Read the answer and worked solution →Photoelectric current falls to zero at which voltage?Read the answer and worked solution →The de Broglie wavelength of a fast electron is comparable with:Read the answer and worked solution →Match the photoelectric termRead the answer and worked solution →

Worked practice questions

For accelerated electron with KE = 100 eV, de Broglie wavelength:

  1. 1.23 Å ≈ 0.123 nm (= 12.27/sqrt(100))
  2. 100 Å
  3. 1 nm
  4. 0.01 nm
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Correct answer: A. 1.23 Å ≈ 0.123 nm (= 12.27/sqrt(100))

λ ≈ 12.27/sqrt(V) Å where V in volts. V = 100: λ ≈ 1.23 Å. Comparable to atomic spacing; gives observable electron diffraction.

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Energy of photon with momentum 3.3 × 10⁻²⁵ kg m/s:

  1. 100 eV
  2. 1 J
  3. E = pc = 3.3 × 10⁻²⁵ × 3 × 10⁸ ≈ 10⁻¹⁶ J ≈ 620 eV
  4. 6.6 × 10⁻³⁴ J
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Correct answer: C. E = pc = 3.3 × 10⁻²⁵ × 3 × 10⁸ ≈ 10⁻¹⁶ J ≈ 620 eV

We have E = pc. That is E = (3.3 × 10⁻²⁵)(3 × 10⁸). So E = 9.9 × 10⁻¹⁷ J ≈ 620 eV. (X-ray energy.) Corresponding λ = h/p = 6.6 × 10⁻³⁴ / 3.3 × 10⁻²⁵ = 2 × 10⁻⁹ m = 2 nm (X-ray range).

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For photoelectric experiment, plot of KE_max vs frequency ν is:

  1. Curve
  2. Parabola
  3. Straight line with slope h, intercept -φ
  4. Horizontal
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Correct answer: C. Straight line with slope h, intercept -φ

KE_max = hν - φ. Linear in ν. Slope = h (Planck constant, universal). Intercept on ν-axis = ν₀ = φ/h (threshold). **Different metals:** same slope, different intercept.

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