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DUAL NATURE OF RADIATION AND MATTER quiz-15

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. The energy of a photon of green light of wavelength 50000 Å is
  •  3.459×10-19 J
  •  3.973×10-19 J
  •  4.132×10-19 J
  •  8453×10-19 J
Solution
E=hc/λ =6.6×10-34×3×108/5000×10-19 =3.973×10-19 J

Q2.Dual nature of radiation is shown by
  •  Diffraction and reflection
  •  Refraction and diffraction
  •  Photoelectric effect alone
  •  Photoelectric effect and diffraction
Solution
{Photoelectric effect →Particle nature Diffraction →Wave nature } Dual nature

Q3. A photon will have less energy, if its
  •   Amplitude is higher
  •  Frequency is higher
  •  Wavelength is longer
  •  Wavelength is shorter
Solution
Energy of a photon E=hv/λ;E is less if λ is longer

Q4. If a voltage applied to an X-ray tube is increased to 1.5 times the minimum wavelength (λmin) of an X-ray continuous spectrum shifts by Δλ=26 pm. The initial voltage applied to the tube is
  •  ≈10 kV
  •  ≈16 kV
  •  ≈50 kV
  •  ≈75 kV
Solution
(b) λmin=hc/eV⇒λ1=hc/(eV1 ) and λ2=hc/(eV2 ) ∴∆λ=λ21=hc/e [1/V2 -1/V1 ].Given V2=1.5 V1 on solving we get V1=16000 volt=16 kV

Q5.A metal plate gets heated when cathode rays strike against it due to
  •  Kinetic energy of cathode rays
  •  Potential energy of cathode rays
  •  Linear velocity of cathode rays
  •  Angular velocity of cathode rays
Solution
When cathode rays strike the metal plate, they transfer their energy to plate  

Q6. An X-ray tube is operated at 50 kV. The minimum wavelength produced is
  •  0.5 Å
  •  0.75 Å
  • 0.25 Å
  •  1 Å
Solution
λmin=12375/(50×103 ) Å=0.247=0.25Å

Q7.If f1,f2 and f3 are the frequencies of corresponding Kα,Kβ and Lα X-rays of an element, then
  •  f1=f2=f3
  •  f1-f2=f3
  •  f2=f1+f3
  •  f22=f1 f3
Solution
For kα emission transition L shell to k- shell For kβ emission transition M shell to k- shell For Lα emission transition M shell to L- shell EM-EK=(EM-EL )+(EL-EK) ⇒hf2=hf3+hf1⇒f2=f1+f3

Q8. De-Broglie wavelength of a body of mass 1 kg moving with velocity of 2000 m/s is
  •  3.32×10-27
  •  1.5×107
  •  0.55×10-22
  •  None of these
Solution
λ=h/mv=(6.6×10-34)/(1×2000)=3.3×10-37 m=3.3×10-27

Q9.Gases begin to conduct electricity at low pressure because
  •  At low pressure, gases turn to plasma
  •  Colliding electrons can acquire higher kinetic energy due to increased mean free path leading to ionization of atoms
  •  Atoms break up into electrons and protons
  •  The electrons in atom can move freely at low pressure
Solution
For ionisation, high energy electrons are required

Q10. The kinetic energy of electron and proton is 10-32 J. Then the relation between their de-Broglie wavelength is
  •  λpe
  •  λpe
  •  λpe
  • λp=2λe
Solution
By using λ=(h/√2mE)  E=10-32 J= Constant for both particles. Hence λ∝1/√m Since mp>me so λpe

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