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. Ultraviolet light of wavelength 300 nm and intensity 1.0 Wm-2 falls on the surface on photoelectric metal. If one percent of incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 cm2 of the surface is nearly
Solution
Energy of each photon =hc/λ = (6.6×10-34×3×108)/(300×10-9) = 6.6×10-19 J Power of source is given by p= intensity × area = 1.0×1.0×10-4 W = 10-4 W No. of photons per second = p/e=10-4/(6.6×10-19) Now number of electron emitted = 1/100×(10-4 /6.6×10-19 ) =1.5×1012 s-1
Energy of each photon =hc/λ = (6.6×10-34×3×108)/(300×10-9) = 6.6×10-19 J Power of source is given by p= intensity × area = 1.0×1.0×10-4 W = 10-4 W No. of photons per second = p/e=10-4/(6.6×10-19) Now number of electron emitted = 1/100×(10-4 /6.6×10-19 ) =1.5×1012 s-1
Q2.Which of the following statement about photon is incorrect?
Solution
Photos move with velocity of light and have energy hv. Therefore, they also exert pressure
Photos move with velocity of light and have energy hv. Therefore, they also exert pressure
Q3. The wavelength of the matter wave is independent of
Solution
Charge
Charge
Q4. For moving ball of cricket, the correct statement about de-Broglie wavelength is
Solution
h/√2mE
h/√2mE
Q5.In above question the energy of the characteristic X-rays given out is
Solution
Less than 40 keV
Less than 40 keV
Q6. Light of frequency v is incident on a substance of threshold frequency v0 (v0<v). The energy of the emitted photoelectron will be
Solution
h(v-v0)
h(v-v0)
Q7.Cathode rays are similar to visible light rays as
Solution
Light consists of photons and cathode rays consists of electrons. However both effect the photographic plate
Light consists of photons and cathode rays consists of electrons. However both effect the photographic plate
Q8.A photo cell is receiving light from a source placed at a distance of 1 m. If the same source is to be placed at a distance of 2 m, then the ejected electron
Solution
Number of ejected electrons ∝ (Intensity) ∝1/(Distance)2 Therefore an increment of distance two times will reduce the number of ejected electrons to 1/4 th of the initial value
Number of ejected electrons ∝ (Intensity) ∝1/(Distance)2 Therefore an increment of distance two times will reduce the number of ejected electrons to 1/4 th of the initial value
Q9.According to Mosely’s law, the frequency of a spectral line in X-ray spectrum varies as
Solution
Square of the atomic number of the element
Square of the atomic number of the element
Q10. Electron with energy 80 keV are incident on the tungsten target of a X-rays tube. Kshell electrons of tungsten have -72.5keV energy. X-rays emitted by the tube contain only
Solution
Since the energy of incident electron, E=80 keV. The minimum wavelength of X-rays produced is λ=hc/E=(6.6×10-34×3×108)/(80×1000×1.6×10-19 ) =1.55×10-10 m=0.155 â„« Since the energy of K-shell electron is -72.5 keV, so the incident electron of energy 80 keV will not only produce continuous spectrum of minimum wavelength 0.155â„« but shell also knock electron of K shell out of atom, resulting emission of characteristics X-rays
Since the energy of incident electron, E=80 keV. The minimum wavelength of X-rays produced is λ=hc/E=(6.6×10-34×3×108)/(80×1000×1.6×10-19 ) =1.55×10-10 m=0.155 â„« Since the energy of K-shell electron is -72.5 keV, so the incident electron of energy 80 keV will not only produce continuous spectrum of minimum wavelength 0.155â„« but shell also knock electron of K shell out of atom, resulting emission of characteristics X-rays