DUAL NATURE OF RADIATION AND MATTER quiz-16
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 charged oil drop of mass 2.5×10-7 kg is in space between the two plates, each of area 2×10-2 m2 of a parallel plate capacitor.
When the upper plate has a charge of 5×10-7 C and the lower plate has an equal negative charge, the oil remains stationary. The charge of the oil drop is [Take g=10 m/s2]
Solution
We know that qE=mg qQ/(ε0 A)=mg or q=(ε0 Amg)/Q =(8.85×10-12×2×10-2×2.5×10-7×10)/(5×10-7 ) C =8.85×10-13 C
We know that qE=mg qQ/(ε0 A)=mg or q=(ε0 Amg)/Q =(8.85×10-12×2×10-2×2.5×10-7×10)/(5×10-7 ) C =8.85×10-13 C
Q2.Light of wavelength 4000â„«is incident on a metal surface. The maximum kinetic energy of emitted photoelectron is 2 eV. What is the work function of the metal surface ?
Solution
According to Einstein’s photoelectric equation the work function of metal is given by ∴ Ï•= hc/λ-KEm = (6.6×10-34×3×108)/(4000×10-10 )-2eV = 4.95×10-19-2 eV = (4.95×10-19)/(1.6×10-19 )-2 eV = 3 eV-2 eV = 1 eV
According to Einstein’s photoelectric equation the work function of metal is given by ∴ Ï•= hc/λ-KEm = (6.6×10-34×3×108)/(4000×10-10 )-2eV = 4.95×10-19-2 eV = (4.95×10-19)/(1.6×10-19 )-2 eV = 3 eV-2 eV = 1 eV
Q3. If λ1 and λ2 are the wavelengths of characteristic X-rays and gamma rays respectively, then the relation between them is
Solution
In general X-rays have larger wavelength than that of gamma rays
In general X-rays have larger wavelength than that of gamma rays
Q4. Who discovered the charge on an electron for the first time?
Solution
Millikan
Millikan
Q5.The wavelength of the photoelectric threshold for silver is λ0. The energy of the electron ejected from the surface of silver by an incident light of wavelength λ(λ<λ0) will be
Solution
Ek=hc/λ-hc/λ0 =hc[(λ0-λ)/(λ0 λ)]
Ek=hc/λ-hc/λ0 =hc[(λ0-λ)/(λ0 λ)]
Q6. When light falls on a metal surface, the maximum kinetic energy of the emitted photo-electrons depends upon
Solution
Kmax=(hv-W0 );v= frequency of incident light
Kmax=(hv-W0 );v= frequency of incident light
Q7.In X-ray experiment Kα,Kβ denotes
Solution
Characteristic
Characteristic
Q8. Light of wavelength 4000 â„« falls on a photosensitive metal and a negative 2V potential stops the emitted electrons. The work function of the material (in eV) is approximately
(h=6.6×10-34 Js,e=1.6×10-19 C,c=3×108 ms-1)
Solution
Energy of incident light E(eV)=12375/4000=3.09 eV Stopping potential is -2V so Kmax=2 eV Hence by using E=W0+Kmax;W0=1.09 eV=1.1 eV
Energy of incident light E(eV)=12375/4000=3.09 eV Stopping potential is -2V so Kmax=2 eV Hence by using E=W0+Kmax;W0=1.09 eV=1.1 eV
Q9.An α-particle of mass 6.4×10-27kg and charge 3.2×10-19 C is situated in a uniform electric field of 1.6×105 Vm-1. The velocity of the particle at the end of 2×10-2 m path when it starts from rest is
Solution
Given, m_α=6.4×10-27)kg qα=3.2×10-19 C, E=1.6×105 Vm-1 Force on α-particle F=qα F=3.2×10-19×1.6×105 =51.2×10-15N Now, acceleration of the particle α=F/mα =(51.2×10-15)/(6.4×10-27 )=0.8×1013 ms-2 ∵Initial velocity, u=0 ∴ v2=2αs =2×8×1012×2×10-2=32×1010 or v=4√2×105 ms-1
Given, m_α=6.4×10-27)kg qα=3.2×10-19 C, E=1.6×105 Vm-1 Force on α-particle F=qα F=3.2×10-19×1.6×105 =51.2×10-15N Now, acceleration of the particle α=F/mα =(51.2×10-15)/(6.4×10-27 )=0.8×1013 ms-2 ∵Initial velocity, u=0 ∴ v2=2αs =2×8×1012×2×10-2=32×1010 or v=4√2×105 ms-1
Q10. When ultraviolet rays are incident on metal plate, then photoelectric effect does not occur. It occurs by the incidence of
Solution
λ(X-ray)<λ(UV-ray)
λ(X-ray)<λ(UV-ray)
