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Published on: 13/05/2022
QB365 provides detailed and simple solution for every Book back Questions in class 12 Physics Subject. It will helps to get more idea about question pattern in every book back questions with solution.
latest Book back QuestionsDownload Tamil Nadu 12th Standard Physics question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Physics Test1.
If the mean wavelength of light from sun is taken as 550 nm and its mean power as 3.8 x 1026 W, then the number of photons emitted per second from the sun is of the order of _____.
1045
1042
1054
1051
2.
Two radiations with photon energies 0.9 eV and 3.3 eV respectively are falling on a metallic surface successively. If the work function of the metal is 0.6 eV, then the ratio of maximum speeds of emitted electrons in the two cases will be _____.
1:4
1:3
1:1
1:9
3.
A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ /2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the material is _____.
\(\frac{hc}{\lambda}\)
\(\frac{2hc}{\lambda}\)
\(\frac{hc}{3\lambda}\)
\(\frac{hc}{2\lambda}\)
4.
When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is \(\frac{V}{4}\). The threshold wavelength for the metallic surface is _____.
4λ
5λ
\(\frac{5}{2}λ\)
3λ
5.
In an electron microscope, the electrons are accelerated by a voltage of 14 kV. If the voltage is changed to 224 kV, then the de Broglie wavelength associated with the electrons would _____.
increase by 2 times
decrease by 2 times
decrease by 4 times
increase by 4 times
1.
\(\mathrm{P} =\frac{\mathrm{n}}{\mathrm{t}} \frac{\mathrm{hc}}{\lambda} \)
\(\frac{\mathrm{n}}{\mathrm{t}} =\frac{\mathrm{P} \lambda}{\mathrm{hc}} \)
\(\frac{\mathrm{n}}{\mathrm{t}} =\frac{3.8 \times 10^{26} \times 550 \times 10^{-9}}{6.6 \times 10^{-3} \times 3 \times 10^8}=1 \times 10^{-15}\)
2.
K.E= hv - Φ
K.E1 = 0.9 - 0.6 = 0.3 eV
K.E2 = 3.3 - 0.6 = 2.7 ev
K.E ∝ v2
\(\frac{0.3}{2.7}=\frac{v^2_1}{v^2_2} \)
\(\frac{v^1}{v^2} =\frac{1}{3}\)
3.
\(\frac{\mathrm{hc}}{\lambda} =\phi+\mathrm{K} . \mathrm{E} .....(1) \)
\(\frac{2 \mathrm{hc}}{\lambda} =\phi+3 \mathrm{~K} . \mathrm{E}......(2)\)
multiply eqn. (1) by 3, we get
\(\frac{3 \mathrm{hc}}{\lambda} =3\phi+3 \mathrm{~K} . \mathrm{E} .....(3)\)
Subtract eqn. (2) from (3), we get
\(\frac{\mathrm{hc}}{\lambda} =2\phi \)
\(\phi=\frac{ \mathrm{hc}}{2\lambda} \)
4.
\(\frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{eV} \) .....(1)
\(\frac{\mathrm{hc}}{2 \lambda}=\phi+\frac{\mathrm{eV}}{4}\) .....(2)
multiply (2) eqn by 4
\(\frac{2 h c}{\lambda}=4 \phi+\mathrm{eV}\) .....(3)
subtract eqn (1) from (3), we get
\(\frac{ h c}{\lambda}=3 \phi \Rightarrow \phi = \frac{ h c}{3\lambda}\)
\(\frac{ h c}{\lambda_o}=\frac{ h c}{3\lambda}\)
⋋o = 3⋋
5.
\(\lambda\propto \frac{1}{\sqrt{V}}\)
\(\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{224\times10^3}}{\sqrt{14\times 10^3}}\)
\(=\sqrt{16}=4\)
\(\lambda_{\mathrm{2}}= \frac{\lambda_1}{4}\)
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