Dual Nature of Matter and Radiation

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De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm) Answer: 2. 0.67 nm Key idea: Bohr's quantization 2π r_n=nλ (electron wave fits the orbit). Radius of the n=2 orbit: r₂=n²a₀=4×0.052=0.208 nm... Matter waves, de Broglie relation 1 answer
A photon and an electron (mass m) have the same energy E. The ratio ≤ft(λ_photon/λ_electron) of their de Broglie wavelengths is: (c is the speed of light) Answer: 3. c√(2m/E) Key idea: λ=h/p , with different momentum-energy relations for photon and electron. Photon: E=pc⇒λ_ph=hc/E . Electron: E... Matter waves, de Broglie relation 1 answer
Which of the following options represent the variation of photoelectric current with property of light shown on the x-axis?A. Photoelectric current vs Intensity of light: straight line through origin risingB. Photoelectric current vs Intensity of lig… Answer: 1. A only Key idea: at a fixed frequency above the threshold, the photoelectric current is directly proportional to the intensity. A... Photoelectric effect, Hertz and Lenard's observations 1 answer
If c is the velocity of light in free space, the correct statements about photon among the following are :A. The energy of a photon is E=h.B. The velocity of a photon is c.C. The momentum of a photon, p=h/c.D. In a photon-electron collision, both tot… Answer: 3. A, B, C and D only Key idea: Go through the properties of a photon. A: E=hν . True. B: Speed is c . True. C: p=hν/c=frac hλ . Tru... Einstein's photoelectric equation, particle nature of light 1 answer
The graph which shows the variation of ≤ft(1/λ^2) and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle) : Answer: 1. graph (1) Key idea: For a free particle E=p²/2m and λ=h/p . So E=h²/2mλ² , which gives frac1λ²=2m/h²E . frac1λ² is directly propo... Matter waves, de Broglie relation 1 answer
The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to: Answer: 3. 1/V The shortest X-ray wavelength occurs when the electron gives all its kinetic energy to one photon: eV=dfrachcλ_min . λ_min=hc... Einstein's photoelectric equation, particle nature of light 1 answer
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? Answer: 2. Cs only A surface emits photoelectrons only if the photon energy is at least its work function. Cs: 2.20 ≥ 2.14 eV, so it emits.... Photoelectric effect, Hertz and Lenard's observations 1 answer
When two monochromatic lights of frequency, and /2 are incident on a photoelectric metal, their stopping potential becomes V_s/2 and V_s respectively. The threshold frequency for this metal is : Answer: 1. 3/2ν Einstein's equation: eV_s=h(ν-ν₀) , so stopping potential is proportional to (ν-ν₀) . Taking the ratio of the two cases: V_s... Einstein's photoelectric equation, particle nature of light 1 answer
The graph which shows the variation of the de Broglie wavelength (λ) of a particle and its associated momentum (p) is : Answer: 1. (graph) de Broglie wavelength is λ=h/p , so λ is inversely proportional to p . The graph of λ against p is therefore a rectangula... Matter waves, de Broglie relation 1 answer
The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3×10^-3 watt will be : (h = 6.6×10^-34 Js) Answer: 3. 10¹⁶ Key idea: number of photons per second N=P/hc/λ=Pλ/hc . N=dfrac3.3×10⁻³×600×10⁻⁹6.6×10⁻³⁴×3×10⁸=dfrac1.98×10⁻⁹1.98×10⁻²⁵=10¹... Einstein's photoelectric equation, particle nature of light 1 answer
An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λ_d, then : Answer: 3. λ = (2mc/h)λ_d² Key idea: with negligible work function, the photoelectron's kinetic energy equals the photon energy. hc/λ=p²/2m⇒... Matter waves, de Broglie relation 1 answer
An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227×10^-2 nm, the potential difference is : Answer: 4. 10⁴ V de Broglie wavelength of an electron accelerated through V volts: λ=1.227/sqrt V nm. Given λ=1.227×10⁻² nm: sqrt V=frac1.22... Matter waves, de Broglie relation 1 answer
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 intensity is doubled ? Answer: 4. zero Photoelectric emission needs frequency ν≥ν₀ , whatever the intensity. Initially ν=1.5ν₀ . Halved, it becomes 0.75ν₀<ν₀ , bel... Photoelectric effect, Hertz and Lenard's observations 1 answer
An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : (m_e = 9×10^-31 kg) Answer: 2. 12.2×10⁻¹² m de Broglie wavelength of an electron accelerated through V : λ=h/√(2meV) . Substituting h=6.63×10⁻³⁴ , m=9×10⁻³¹ , e... Matter waves, de Broglie relation 1 answer
When the light of frequency 2_0 (where _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_0, the maximum velocity of electrons emi… Answer: 1. 1 : 2 Einstein's equation: K_max=hν-hν₀ . At 2ν₀ : K₁=hν₀ . At 5ν₀ : K₂=4hν₀ . Since K=1/2mv² , v∝sqrt K : v₁/v₂=√(1/4)=1/2 Ratio... Einstein's photoelectric equation, particle nature of light 1 answer
An electron of mass m with an initial velocity V=V_0 i (V_0>0) enters an electric field E=-E_0 i (E_0 = constant >0) at t = 0. If λ_0 is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is Answer: 1. λ₀/(1+eE₀/mV₀t) The electron has charge -e and the field is -E₀hat i , so the force is +eE₀hat i . It accelerates along its motio... Matter waves, de Broglie relation 1 answer
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) Answer: Dropped in the official answer key: full marks to every candidate Key idea: 1/2mv²=hc/λ-hc/λ₀ . hc=4.14×10⁻¹⁵×3×10⁸=1242 eV nm. Phot... Einstein's photoelectric equation, particle nature of light 1 answer
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is : Answer: 1. h/√(3mkT) Key idea: a thermal neutron has average kinetic energy 3/2kT , and λ=h/p=h/√(2mK) . λ=h/√(2m·3/2kT)=h/√(3mkT) . Matter waves, de Broglie relation 1 answer
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… Answer: 3. -3 V Einstein's equation: K_max=hν-φ . 5 eV photon, K_max=2 eV, so φ=3 eV. 6 eV photon: K_max=6-3=3 eV. To stop all electrons the... Einstein's photoelectric equation, particle nature of light 1 answer
Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ_0) of the emitted X-ray is Answer: 4. λ₀=2mcλ²/h The cutoff wavelength corresponds to the whole kinetic energy K of the electron turning into one photon: λ₀=hc/K . de... Matter waves, de Broglie relation 1 answer
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 i… Answer: 4. 3λ Key idea: eV=hc/λ-hc/λ₀ . eV=hc/λ-φ and eV/4=hc/2λ-φ . Subtracting: 3/4eV=hc/2λ⇒ eV=2hc/3λ . φ=hc/λ-2hc/3λ=hc/3λ , so λ₀=3λ . Einstein's photoelectric equation, particle nature of light 1 answer
An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is : (c being velocity of light) Answer: 1. 1/c(E/2m)^(1/2) Key idea: use λ=h/p for the electron and λ=hc/E for the photon. Electron: p=√(2mE) , so λ_e=h/√(2mE) . Photon: λ_... Matter waves, de Broglie relation 1 answer