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Published on: 01/06/2021
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Questions + Answers key
Take MCQ Physics Test1.
At the given point of time, the earth receives energy from sun at 4 cal cm–2min–1. Determine the number of photons received on the surface of the Earth per cm2 per minute. (Given: Mean wavelength of sun light = 5500 Å )
2.
Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for the photoelectric emission experiment.
3.
When a 6000Å light falls on the cathode of a photo cell, photoemission takes place. If a potential of 0.8 V is required to stop emission of electron, then determine the
(i) frequency of the light
(ii) energy of the incident photon
(iii) work function of the cathode material
(iv) threshold frequency and
(v) net energy of the electron after it leaves the surface.
4.
What should be the velocity of the electron so that its momentum equals that of 4000 Å wavelength photon.
5.
A 150 W lamp emits light of mean wavelength of 5500 Å. If the efficiency is 12%, find out the number of photons emitted by the lamp in one second.
1.
\(P=4 \ \mathrm{cal} \mathrm{} \mathrm{cm}^{-2} \mathrm{~min}^{-1}=4 \times 4.2=16.8 \mathrm{~J} \mathrm{~cm}^{-2} \mathrm{~min}^{-1} \)
\(E=\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{5500 \times 10^{-10}}=3.6 \times 10^{-19} \mathrm{~J} \)
\(n=\frac{E}{hv}=\frac{16.8}{3.6 \times 10^{-19}}=4.67 \times 10^{19} \)
\(n=4.67 \times 10^{19} \) per cm2 per minute.
2.
V = 81 V
∴ K = eV = 1.6 x 10-19 x 81 = 1.296 x 10-17 = 1.3 x 10-17 J
\({ V }=\sqrt \frac {2K}{m}\sqrt { \cfrac { 2\times 1.3\times { 10 }^{ -17 } }{ 9.1\times { 10 }^{ -31 } } } =5.345 \times 10^6 ms^{-1} \)
v = 5.345 x 106 m s-1, K = 1.3 x 10-17 J
3.
\(\lambda=6000Å=6000 \times 10^{-10} \mathrm{~m} ; \mathrm{V}=0.8 \mathrm{v} \)
\(\mathrm{k} \cdot \mathrm{E}=\mathrm{hv}-\phi \)
\(\mathrm{eV}_o=\mathrm{hv}-\phi=\frac{\mathrm{hc}}{\lambda}-\phi \)
\((i) v=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6000 \times 10^{-10}}=5 \times 10^{14} \mathrm{~Hz} \)
\((ii)\ \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-10}}=3.313 \times 10^{-19} J\)
\(\mathrm{E}=\frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}}=2.07 \mathrm{eV} \)
\((iii) \ \mathrm{E}=\mathrm{hv}-\mathrm{W}\Rightarrow\mathrm{W}=\mathrm{h} v-\mathrm{E} \)
\(\mathrm{E}=\mathrm{eV_o}=1.6 \times 10^{-19} \times 0.8=1.2 8 \times10^{-19}J\)
\(\mathrm{hv}=6.626 \times 10^{-34} \times 5 \times 10^{14}=3.313 \times 10^{-19}J \)
\(\mathrm{~W}=\frac{(3.313-1.28) \times 10^{-19}}{1.6 \times 10^{-19}}=1.270 \mathrm{eV} \)
W = 1.27 eV
\((iv) \ \mathrm{W}=\mathrm{h} \mathrm{v}_{0} \)
\(v_{0}=\frac{W}{h}=\frac{2.033 \times 10^{-19}}{6.626 \times 10^{-34}}=3.07 \times 10^{14} \mathrm{~Hz} \)
\((v)\ \mathrm{E}=\mathrm{eV}_o=\frac{0.8 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=\mathbf{0 . 8} \mathrm{eV}\)
4.
\(\lambda=4000 \stackrel{o}A=4000 \times 10^{-10} \mathrm{~m} \)
\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\Rightarrow \mathrm{v }=\frac{\mathrm{h}}{\mathrm{m\lambda}}=\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 4000 \times 10^{10}}=1820 \mathrm{~ms}^{-1} \)
5.
λ = 5500 x 10-10 m; P =150 W; Efficiency =12%
\({ E }=\cfrac { hc }{ { \lambda } } =\cfrac { 6.626\times { 10 }^{ -34 }\times 3 \times { 10 }^{ 8 } }{ 5500\times { 10 }^{ -10 } }=3.614 \times 10^{-19} J \)
\(n=\frac{E}{hv}=\cfrac { 150 }{ 3.614\times { 10 }^{ -19 } } =4.15 \times 10^{20}\times\frac{12}{100}\)
n = 4.98 x 1019s-1
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Biology

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Commerce

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