DUAL NATURE OF RADIATION AND MATTER QUIZ-11
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. In a discharge tube at 0.02 mm, there is a formation of
Solution
CDS
CDS
Q2.Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. the threshold frequency of the material is
Solution
n→2-1 E=10.2 eV kE=E-Ï• Q=10.20-3.57 hv0=6.63 eV v0=(6.63×1.6×10-19)/(6.67×10-34 )=1.6×1015
n→2-1 E=10.2 eV kE=E-Ï• Q=10.20-3.57 hv0=6.63 eV v0=(6.63×1.6×10-19)/(6.67×10-34 )=1.6×1015
Q3. The curve between current (i)and potential difference(V) for a photo cell will be
Solution
In photocell, at a particular negative potential (stopping potential V0) of anode, photoelectric current is zero, as the potential difference between cathode and anode increases current through the circuit increases but after some time constant current (saturation current) flows through the circuit even if potential difference still increases
In photocell, at a particular negative potential (stopping potential V0) of anode, photoelectric current is zero, as the potential difference between cathode and anode increases current through the circuit increases but after some time constant current (saturation current) flows through the circuit even if potential difference still increases
Q4. Penetrating power of X-rays can be increased by
Solution
With the increase in potential difference between anode and cathode energy of striking electrons increases which in turn increses the energy (penetration power) of X-rays
With the increase in potential difference between anode and cathode energy of striking electrons increases which in turn increses the energy (penetration power) of X-rays
Q5.In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is 2×10-7 and stopping potential is 2.5 V. The threshold frequency of the metal (in Hz ) approximately (charge on electron e=1.6×10-19 C, Planck’s constant h=6.6×10-34J-s)
Solution
eV0=hv-hv0 ∴ Threshold frequency, v0=v-(eV0)/h = c/λ-(eV0)/h ∴ v0=(3×108)/(2×10-7 )-(1.6×10-19×2.5)/(6.6×10-34 ) = 9.0×1014 Hz
eV0=hv-hv0 ∴ Threshold frequency, v0=v-(eV0)/h = c/λ-(eV0)/h ∴ v0=(3×108)/(2×10-7 )-(1.6×10-19×2.5)/(6.6×10-34 ) = 9.0×1014 Hz
Q6. The maximum wavelength of radiation that can produce photoelectric effect in certain metal is 200 nm. The maximum kinetic energy acquired by electron due to radiation of wavelength 100 nm will be
Solution
Here, λ0=200nm; λ=100nm; hc/e = 1240eV nm maximum KE =hc/λe-hc/(λ0 e)(in eV) =hc/e (1/λ-1/λ0 ) =1240(1/100-1/200) =6.2 eV
Here, λ0=200nm; λ=100nm; hc/e = 1240eV nm maximum KE =hc/λe-hc/(λ0 e)(in eV) =hc/e (1/λ-1/λ0 ) =1240(1/100-1/200) =6.2 eV
Q7.From the following, what charges can be present on oil drops in Millikan’s experiment
(Here e is the electronic charge)
Solution
In Milikan’s experiment, the charges present on the oil drops are the integral multiples of electronic charge, so 2e and 10e(1.6×10-18 C) charges are present
In Milikan’s experiment, the charges present on the oil drops are the integral multiples of electronic charge, so 2e and 10e(1.6×10-18 C) charges are present
Q8.Which of the following metal thermionically emits an electron at a relatively lowest temperature among them
Solution
Among the given metals, aluminium thermionically emits an electron at a relatively lowest temperature
Among the given metals, aluminium thermionically emits an electron at a relatively lowest temperature
Q9.The temperature at which protonsin proton gas would have enough energy to overcome Coulomb barrier of 4.14×10-14 J is (Boltzmann constant
=1.38×10-23 JK-1)
Solution
Given k=1.38×10-23 JK-1 The energy of proton gas =4.14×10-14 J E=3/2 kT 4.14×10-14=3/2×1.38×10-23×T T=2×109 K.
Given k=1.38×10-23 JK-1 The energy of proton gas =4.14×10-14 J E=3/2 kT 4.14×10-14=3/2×1.38×10-23×T T=2×109 K.
Q10. The longest wavelength that can be analysed by a sodium chloride crystal of spacing d=2.82 A° in the second order is
Solution
Use Bragg’s X-ray diffraction Law nλ=2d sinθ ∴λ=(2d sinθ)/n For longest wavelength take sin θ=1 ∴λ=(2×2.82[â„«])/2[∵n=2 for second order] =2.82 â„«
Use Bragg’s X-ray diffraction Law nλ=2d sinθ ∴λ=(2d sinθ)/n For longest wavelength take sin θ=1 ∴λ=(2×2.82[â„«])/2[∵n=2 for second order] =2.82 â„«
