JEE Physics Dual Nature of Radiation and Matter MCQs Set 23

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Chapter-wise Objective Questions: Dual Nature Of Radiation and Matter

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Multiple Choice Questions

Question 1: What are the frequency and energy of a photon of wavelength 6000 \( \mathring{A} \)? (h = 6.6 x 10\(^{-34}\) J s, c = 3 x 10\(^8\) m/s)
(a) \( 15\times10^{4}\ \text{Hz and } 3.3\times10^{-10}\ \text{J} \)
(b) \( 6\times10^{10}\ \text{Hz and } 4.4\times10^{-19}\ \text{J} \)
(c) \( 5\times10^{14}\ \text{Hz and } 3.3\times10^{-19}\ \text{J} \)
(d) \( 5\times10^{12}\ \text{Hz and } 3.3\times10^{-19}\ \text{J} \)
Show Answer & Explanation

Answer: (c) \( 5\times10^{14}\ \text{Hz and } 3.3\times10^{-19}\ \text{J} \)

Explanation:
1. \( \nu=c/\lambda=3\times10^8/(6000\times10^{-10})=5\times10^{14} \) Hz.
2. \( E=h\nu=6.6\times10^{-34}\times5\times10^{14}=3.3\times10^{-19} \) J.

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Question 2: The work function of sodium is 2.3 eV. The maximum wavelength of light that will cause photoelectrons to be emitted from sodium is
(a) 5400 \( \mathring{A} \)
(b) 6400 \( \mathring{A} \)
(c) 5000 \( \mathring{A} \)
(d) 5200 \( \mathring{A} \)
Show Answer & Explanation

Answer: (a) 5400 \( \mathring{A} \)

Explanation:
1. The threshold condition is \( hc/\lambda_{max}=\phi_0 \).
2. \( \lambda_{max}=\dfrac{hc}{\phi_0}=\dfrac{1240\ \text{eV nm}}{2.3\ \text{eV}}\approx540 \) nm = 5400 Å.

Teacher's Note:
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Question 3: In a photoelectric experiment, for incident light of wavelength \( 1.98\times10^{-7} \) m the stopping potential is 2.5 V. The maximum kinetic energy of the photoelectrons, the work function and the threshold frequency are respectively
(a) \( 6.25\ \text{eV},\ 3.75\ \text{eV},\ 9.1\times10^{14}\ \text{Hz} \)
(b) \( 2.5\ \text{eV},\ 3.75\ \text{eV},\ 9.1\times10^{14}\ \text{Hz} \)
(c) \( 3.75\ \text{eV},\ 6.25\ \text{eV},\ 9.1\times10^{14}\ \text{Hz} \)
(d) \( 4.75\ \text{eV},\ 6.25\ \text{eV},\ 8.1\times10^{14}\ \text{Hz} \)
Show Answer & Explanation

Answer: (b) \( 2.5\ \text{eV},\ 3.75\ \text{eV},\ 9.1\times10^{14}\ \text{Hz} \)

Explanation:
1. Maximum kinetic energy \( =eV_0=2.5 \) eV.
2. Photon energy \( =1240/198=6.25 \) eV, so \( \phi_0=6.25-2.5=3.75 \) eV.
3. \( \nu_0=\phi_0/h=3.75\times1.6\times10^{-19}/6.6\times10^{-34}=9.1\times10^{14} \) Hz.

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Question 4: The ratio of the de-Broglie wavelength of a proton to that of an alpha particle accelerated through the same potential difference is
(a) \( 3\sqrt2 \)
(b) \( 2\sqrt3 \)
(c) \( 2\sqrt5 \)
(d) \( 2\sqrt2 \)
Show Answer & Explanation

Answer: (d) \( 2\sqrt2 \)

Explanation:
1. \( \lambda=h/\sqrt{2mqV} \).
2. \( \dfrac{\lambda_p}{\lambda_\alpha}=\sqrt{\dfrac{m_\alpha q_\alpha}{m_pq_p}}=\sqrt{4\times2}=2\sqrt2 \).

Teacher's Note:
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Question 5: A parallel beam of light is incident normally on a plane surface that absorbs 40% of the light and reflects the rest. If the incident beam carries 60 W of power, the force exerted on the surface is
(a) \( 3.2\times10^{-7}\ \text{N} \)
(b) \( 3.2\times10^{-8}\ \text{N} \)
(c) \( 5.12\times10^{-7}\ \text{N} \)
(d) \( 5.12\times10^{-8}\ \text{N} \)
Show Answer & Explanation

Answer: (a) \( 3.2\times10^{-7}\ \text{N} \)

Explanation:
1. Incident momentum per second \( =P/c=60/(3\times10^8)=2\times10^{-7} \) N.
2. Absorbed part gives \( 0.4\times2\times10^{-7} \); reflected part gives \( 2\times0.6\times2\times10^{-7} \).
3. \( F=2\times10^{-7}(0.4+1.2)=3.2\times10^{-7} \) N.

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Question 6: The energy of a photon whose momentum is \( 3.3\times10^{-13} \) kg m/s is (c = \( 3\times10^8 \) m/s)
(a) \( 7.3\times10^{4}\ \text{J} \)
(b) \( 1.3\times10^{5}\ \text{J} \)
(c) \( 8.1\times10^{3}\ \text{J} \)
(d) \( 9.9\times10^{-5}\ \text{J} \)
Show Answer & Explanation

Answer: (d) \( 9.9\times10^{-5}\ \text{J} \)

Explanation:
1. For a photon \( E=pc \).
2. \( E=3.3\times10^{-13}\times3\times10^8=9.9\times10^{-5} \) J.

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Question 7: A particle of mass 1 mg has the same de-Broglie wavelength as an electron moving with a velocity of \( 3\times10^6 \) m/s. The velocity of the particle is (\( m_e=9.1\times10^{-31} \) kg)
(a) \( 3\times10^{-31}\ \text{m/s} \)
(b) \( 2.7\times10^{-18}\ \text{m/s} \)
(c) \( 2.7\times10^{-21}\ \text{m/s} \)
(d) \( 9\times10^{-2}\ \text{m/s} \)
Show Answer & Explanation

Answer: (b) \( 2.7\times10^{-18}\ \text{m/s} \)

Explanation:
1. Equal wavelengths mean equal momenta: \( mv=m_ev_e \).
2. \( v=\dfrac{9.1\times10^{-31}\times3\times10^6}{10^{-6}}=2.73\times10^{-18} \) m/s.

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Question 8: The energy that must be added to an electron to reduce its de-Broglie wavelength from \( 10^{-10} \) m to \( 0.5\times10^{-10} \) m is
(a) three times the initial energy
(b) four times the initial energy
(c) equal to the initial energy
(d) twice the initial energy
Show Answer & Explanation

Answer: (a) three times the initial energy

Explanation:
1. \( \lambda\propto1/\sqrt E \), so halving \( \lambda \) makes \( E \) four times larger.
2. Energy to be added \( =4E-E=3E \).

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Question 9: The wavelength of a photon needed to remove a proton from a nucleus, in which it is bound with an energy of 1 MeV, is nearly
(a) 1.2 nm
(b) \( 1.2\times10^{-6} \) nm
(c) \( 1.2\times10^{-3} \) nm
(d) 12 nm
Show Answer & Explanation

Answer: (c) \( 1.2\times10^{-3} \) nm

Explanation:
1. \( \lambda=\dfrac{hc}{E}=\dfrac{1240\ \text{eV nm}}{10^6\ \text{eV}} \).
2. \( \lambda=1.24\times10^{-3} \) nm.

Teacher's Note:
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Question 10: Two sources each emit light with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light of wavelength 500 nm. The ratio of the number of X-ray photons to the number of visible-light photons emitted per second is
(a) 1 : 500
(b) 1 : 250
(c) 1 : 20
(d) 100 : 1
Show Answer & Explanation

Answer: (a) 1 : 500

Explanation:
1. Number of photons per second \( =P/(hc/\lambda)=P\lambda/hc \), proportional to \( \lambda \) for equal power.
2. Ratio \( =1:500 \).

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Question 11: A metal is exposed to light of wavelength 600 nm. The maximum kinetic energy of the electrons doubles when light of wavelength 400 nm is used. The work function in eV is
(a) 1.02 eV
(b) 1.50 eV
(c) 1.94 eV
(d) 2.76 eV
Show Answer & Explanation

Answer: (a) 1.02 eV

Explanation:
1. \( K_1=\dfrac{1240}{600}-\phi=2.067-\phi \) and \( K_2=\dfrac{1240}{400}-\phi=3.1-\phi \).
2. \( K_2=2K_1 \) gives \( 3.1-\phi=4.133-2\phi \), so \( \phi=1.03 \) eV, about 1.02 eV.

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Question 12: An important spectral emission line of hydrogen has a wavelength of 21 cm. The corresponding photon energy is (h = \( 6.62\times10^{-34} \) J s, c = \( 3\times10^8 \) m/s)
(a) \( 5.9\times10^{-8}\ \text{eV} \)
(b) \( 5.9\times10^{-4}\ \text{eV} \)
(c) \( 11.8\times10^{-6}\ \text{eV} \)
(d) \( 5.9\times10^{-6}\ \text{eV} \)
Show Answer & Explanation

Answer: (d) \( 5.9\times10^{-6}\ \text{eV} \)

Explanation:
1. \( E=hc/\lambda=6.62\times10^{-34}\times3\times10^8/0.21=9.46\times10^{-25} \) J.
2. \( E=9.46\times10^{-25}/1.6\times10^{-19}=5.9\times10^{-6} \) eV.

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Question 13: The energy of a photon of green light of wavelength 5000 \( \mathring{A} \) is (h = \( 6.63\times10^{-34} \) J s)
(a) \( 3.973\times10^{-19}\ \text{J} \)
(b) \( 3.459\times10^{-19}\ \text{J} \)
(c) \( 4.132\times10^{-19}\ \text{J} \)
(d) \( 8.453\times10^{-19}\ \text{J} \)
Show Answer & Explanation

Answer: (a) \( 3.973\times10^{-19}\ \text{J} \)

Explanation:
1. \( E=\dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3\times10^8}{5\times10^{-7}} \).
2. \( E=3.98\times10^{-19} \) J, about 2.5 eV.

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Question 14: A sodium vapour lamp consumes 10 W of power and 90% of the consumed energy is converted into light of wavelength 590 nm. The number of photons emitted per second is about (h = \( 6.63\times10^{-34} \) J s)
(a) \( 2.67\times10^{18} \)
(b) \( 2.67\times10^{19} \)
(c) \( 2.67\times10^{20} \)
(d) \( 2.67\times10^{17} \)
Show Answer & Explanation

Answer: (b) \( 2.67\times10^{19} \)

Explanation:
1. Power converted to light \( =0.9\times10=9 \) W.
2. Energy of one photon \( =hc/\lambda=6.63\times10^{-34}\times3\times10^8/590\times10^{-9}=3.37\times10^{-19} \) J.
3. \( N=9/3.37\times10^{-19}=2.67\times10^{19} \) per second.

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Question 15: If the energy of photons of wavelength 6000 \( \mathring{A} \) is \( 3.2\times10^{-19} \) J, the photon energy for a wavelength of 4000 \( \mathring{A} \) is
(a) \( 1.11\times10^{-19}\ \text{J} \)
(b) \( 2.22\times10^{-19}\ \text{J} \)
(c) \( 4.40\times10^{-19}\ \text{J} \)
(d) \( 4.80\times10^{-19}\ \text{J} \)
Show Answer & Explanation

Answer: (d) \( 4.80\times10^{-19}\ \text{J} \)

Explanation:
1. \( E\propto1/\lambda \).
2. \( E'=3.2\times10^{-19}\times\dfrac{6000}{4000}=4.8\times10^{-19} \) J.

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Question 16: A radio transmitter operates at a frequency of 880 kHz and a power of 10 kW. The number of photons emitted per second is about (h = \( 6.63\times10^{-34} \) J s)
(a) \( 1.7\times10^{31} \)
(b) \( 1.3\times10^{25} \)
(c) \( 1.3\times10^{37} \)
(d) \( 1.3\times10^{45} \)
Show Answer & Explanation

Answer: (a) \( 1.7\times10^{31} \)

Explanation:
1. Energy of one photon \( =h\nu=6.63\times10^{-34}\times8.8\times10^5=5.83\times10^{-28} \) J.
2. \( N=10^4/5.83\times10^{-28}=1.7\times10^{31} \) per second.

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Question 17: The photoelectric threshold wavelength of tungsten is 2300 \( \mathring{A} \). The maximum energy of the electrons ejected by ultraviolet light of wavelength 1800 \( \mathring{A} \) is
(a) 1.5 eV
(b) 0.15 eV
(c) 15 eV
(d) 150 eV
Show Answer & Explanation

Answer: (a) 1.5 eV

Explanation:
1. \( K_{max}=hc\left(\dfrac1\lambda-\dfrac1{\lambda_0}\right)=1240\left(\dfrac1{180}-\dfrac1{230}\right) \) eV.
2. \( =6.89-5.39=1.5 \) eV.

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Question 18: Ultraviolet radiation of energy 6.2 eV falls on an aluminium surface of work function 4.2 eV. The kinetic energy of the fastest electron emitted, in joule, is approximately
(a) \( 3\times10^{-21} \)
(b) \( 3.2\times10^{-19} \)
(c) \( 3\times10^{-17} \)
(d) \( 3\times10^{-15} \)
Show Answer & Explanation

Answer: (b) \( 3.2\times10^{-19} \)

Explanation:
1. \( K_{max}=6.2-4.2=2.0 \) eV.
2. \( 2.0\times1.6\times10^{-19}=3.2\times10^{-19} \) J.

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Question 19: Radiations whose photon energies are twice and ten times the work function of a metal are incident successively on the metal surface. The ratio of the maximum velocities of the photoelectrons in the two cases is
(a) 1 : 3
(b) 1 : 2
(c) 1 : 4
(d) 1 : 1
Show Answer & Explanation

Answer: (a) 1 : 3

Explanation:
1. \( K_1=2\phi-\phi=\phi \) and \( K_2=10\phi-\phi=9\phi \).
2. \( v\propto\sqrt K \), so \( v_1:v_2=1:3 \).

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Question 20: For a certain metal, when \( \nu=2\nu_0 \) the electrons come out with a maximum velocity of \( 4\times10^6 \) m/s. When \( \nu=5\nu_0 \), the maximum velocity of the photoelectrons will be
(a) \( 2\times10^{7}\ \text{m/s} \)
(b) \( 2\times10^{6}\ \text{m/s} \)
(c) \( 8\times10^{6}\ \text{m/s} \)
(d) \( 8\times10^{5}\ \text{m/s} \)
Show Answer & Explanation

Answer: (c) \( 8\times10^{6}\ \text{m/s} \)

Explanation:
1. \( K=h\nu-h\nu_0 \): at \( 2\nu_0 \), \( K_1=h\nu_0 \); at \( 5\nu_0 \), \( K_2=4h\nu_0 \).
2. \( v_2/v_1=\sqrt{K_2/K_1}=2 \), so \( v_2=8\times10^6 \) m/s.

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