QUIZ-18 DUAL NATURE OF RADIATION AND MATTER

DUAL NATURE OF RADIATION AND MATTER quiz-18

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. A positively charged particle enters a magnetic field of value with a velocity

. The particle will move along

•  + X axis
•  -X axis
•  + Z axis
•  -Z axis

Solution
Force on the charged particle in magnetic field is 

 

Q2.In a photoelectric effect measurement, the stopping potential for a given metal is found to be V₀ volt when radiation of wavelength λ₀ is used. If radiation of wavelength 2λ₀ is used with the same metal then the stopping potential (in volt ) will be

•  V₀/2
•  2V₀
•   V₀+hc/(2eλ₀ )
•   V₀-hc/(2eλ₀ )

Solution
From Einstein’s photoelectric equation eV₀=hc/λ₀ -W₀ eV^'=hc/(2λ₀ )-W₀ Subtracting e(V₀-V^' )=hc/λ₀ [1-1/2]=hc/(2λ₀ ) or V^'= V₀-hc/(2eλ₀ )

Q3. In Thomson mass spectrograph, singly and doubly ionised particles from similar parabola corresponding to magnetic fields of 0.8 T and 1.2 T for a constant electric field. The ratio of masses f ionised particles will be

•  3 : 8
•  2 : 9
•  8 : 3
•  9 : 2

Solution
For similar parabola; y²=(B² lD)/E q/m x, will be same for two particles. It means (B² q)/m= a constant for these two particles. ∴ m₁/m₂ =(B₁² q₁)/(B₂² q₂ )=(0.8/1.2)²×e/2e=2/9

Q4. When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectron and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively

•  N and 2T
•  2N and T
•  2N and 2T
•   N and T

Solution
Number of photoelectrons ∝ Intensity Maximum kinetic energy is independent of intensity

Q5.The ratio of specific charge of an α-particle to that of a proton is

•  2 :1
•  1 :1
•  1 :2
•  1 :3

Solution
Specific charge =q/m; Ratio =(q/m)_α/(q/m)_p =q_α/q_p ×m_p/m_α =1/2  

Q6. Cathode rays are produced when the pressure is of the order of

•  2 cm of Hg
•  0.1 cm of Hg
• 0.01 mm of Hg
•  1 μm ofHg

Solution
0.01 mm of Hg

Q7.Which of the following is not the property of a cathode ray

•  It casts shadow
•  It produces heating effect
•  It produces flurosence
•  It does not deflect in electric field

Solution
Cathode rays are steam of negatively charged particles, so they deflect in electric field

Q8. Maximum velocity of photoelectron emitted is 4.8 ms^-1. The e/m ratio of electron is 1.76×10¹¹ Ckg^-1, then stopping potential is given by

•  5×10^-10 JC^-1
•  3×10^-7 JC^-1
•  7×10^-11 JC^-1
•  2.5×10^-2 JC^-1

Solution
eV_s=1/2 mv_m² orV_s=(mv_m²)/2e=(v_m²)/(2(e/m)) =(4.8)²/(2×17.6×10¹¹ )=7×10¹¹ JC^-1

Q9.When green light is incident on the surface of metal, it emits photo-electrons but there is no such emission with yellow colour light. Which one of the colours can produce emission of photo-electrons

•  Orange
•  Red
•  Indigo
•  None of the above

Solution
Wave length of green light is threshold wave length. Hence for emission of electron, wave length of Indigo light < wavelength of green light

Q10. An α-particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de-Broglie wavelengths are λ_α and λ_p respectively. The ratio λ_p/λ_α , to the nearest integer, is

•  3
•  4
•  2
•  4.5

Solution
λ=h/p=h/√2qVm or λ∝1/√qm λ_p/λ_∝ =√(q_α/q_p .m_α/m_p )=√((2)(4)/(1)(1)) =2.828 The nearest integer is 3.