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At $ 10 ^ \circ C $ temperature, de-Broglie wave lengthof atomis $ 0.4 A ^\circ $ . If temperature of atom is increased by $ 30 ^\circ C $ , what will be change in de-Broglie wavelength of atom ?

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Explanation

$ T_1 = 10 +273 = 283 K $ $ T_2 = 40 +273 = 313 K $ $ \lambda _1 = 0.4 A$ $ \lambda = { h \over \sqrt {2mE} }$ $ E = {3 /2 } KT $ $ \lambda = { h \over \sqrt { 3mKT}}$ $ \therefore \lambda \alpha { 1 \over \sqrt T} $ $ \therefore { \lambda_1 \over \lambda_2} = \sqrt { T_1 \over T_2}$ $ \therefore {\lambda_2 \over 0.4 \times 10^{-10}} = \sqrt {283 \over313}$ $ \therefore {\lambda_2 \over 0.4 \times 10^{-10} }= 0.951 $ $ \therefore \lambda_2 = 0.951 \times 0.4 \times 10^{-10}$ $ \therefore \lambda _2 = 0.38 A $ $ = \lambda_2 - \lambda_1 $ $ = 0.38 A ^\circ - 0.4 A ^\circ = - 0.02 A ^\circ $ $ = 2 \times 10^ {-2} A ^\circ decreases $

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Wavelength of incident radiation on photo sensitive surface is changed from $ 4000 A ^\circ $ to $3000 A ^ \circ $ , so change in stopping potential will be.......

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Explanation

$ \lambda_1 = 4000 A^ \circ = 4 \times 10^{-7}$ $ \lambda _2 = 3600 A^ \circ = 3.6 \times 10^{-7} m $ $V_0 e = { hc \over \lambda } - \phi .....(1)$ $ \therefore V_{01}e = { hc \over \lambda_2} -\phi.....(2)$ $ V_{02}e - V_{01}e = hc [{1 \over \lambda_2} - {1 \over \lambda_2} ]$ $ \therefore V_{02} -V_{01} = { hc \over e} [{ 1 \over \lambda_2 } - {1 \over \lambda _1} ] $ $ = { 6.62 \times 10^{-34} \times 3 \times 10^8 \over 1.6 \times 10^{-19}}[{10^7 \over 3.6} - {10^7 \over 4 } ]$ $ = {6.62 \times 3 \over 1.6 } [{4-3.6 \over 4 \times 3.6 }] $ $ \therefore V_{02} -V_{01} = 0.345 V$

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Wavelength of incident radiation on photo sensitive surface is $ 4000 A ^ \circ $ If wavelength ofthis light is $ 3600 A ^ \circ $ , what will be change in kinetic energy of emitted photo electron ?$ ( h = 6.625 \times 10^{-34} J.s, c = 3 \times 10^8 ms^{-1}, 1ev = 1.6 \times 10^{-19} J)$

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Explanation

$\lambda_1 = 4000 A^\circ = 4 \times 10^ -7 m $ $ \lambda_2 = 3000 A^\circ = 3 \times 10 ^ {-7} m$ $ \triangle = ? $ $ E = { hc \over \lambda } -\phi $ $ E_1 = { hc \over \lambda_1} -\phi ...(1) $ $ E_2 = { hc \over \lambda_2} - \phi...(2) $ $ \therefore E_2 -E_1 = hc ({1 \over \lambda_2 } - { 1 \over \lambda_1} )$ $ = 6.625 \times 10^{-34} \times 3 \times 10^{8} [{10^7 \over 3} - {10^7 \over 4}] = 19.956 \times 10^{-19} [{1 \over 12}]$ $ \therefore \triangle E = 1.04 eV$

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A hollow metallic cuboidal has mass 10 kg and length 30 cm. At what speed this cuboidal is moved in X-direction so that its de-Broglie wavelength exactly trapped in the cuboidal ?

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Explanation

$ m = 10 kg , \triangle x = 30 \times 10^{-2} m $ $ h = \triangle x = {h \over mv}$ $ \therefore v = { h \over m\triangle x} = { 6.625 \times 10^{-34} \over 10 \times 30 \times 10 ^ {-2} } $ $ \therefore v = 2.2 \times 10^ {-34} m/s $

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If we take accelerating voltage V = 50 V, electric charge of electron$ e = 1.6 \times 10 ^ {-19} C$ and mass of electron $m = 9.1 \times 10^{-31} kg$ find the wavelength of concerned electrion.

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Explanation

$ v = 50 V $ $ e = 1.6 \times 10^ {-19} c $ $ m_e = 9.1 \times 10^{-31} kg $ $ h = 6.62 \times 10^{-34} J.s $ $ \lambda = { h \over \sqrt {2meV}}$ $ = {6.62 \times 10^{-34} \over \sqrt{2 \times 9.1 \times 10^{-34} \times 1.6 \times 10^{-19} \times 50}}$ $ \lambda = 1.735 A^ \circ $

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What will be energyin eV of photons of $\lambda$- rays having length $ 0.1A ^ \circ $coming out of excited nucleus of radium ? $( c = 3 \times 10 ^8,h = 6.625 \times 10^{-34} m/s )$

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Explanation

$ \lambda = 0.1 A^ \circ $ $ E = { hc \over \lambda } = { 6.625 \times 10 ^ {-34} \times 3 \times 10 ^ {8}}$ $ = 19.875 \times 10 ^ {-15} J $ $ 1eV = 1.6 \times 10^ {-19 } J$ $ \therefore E ={ 19.875 \times 10^{-15} \over 1.6 \times 10^ {-19} }$ $= 12.42 \times 10^4 eV$

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How many photons of red coloured light having wavelength $ 8000 A ^ \circ $ will have same energyas one photon of violet coloured light of wavelength $ 4000 A ^ \circ $ ?

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Explanation

$ E_1 = {hc \over \lambda_1} $ $ E_2 ={ nhc \over \lambda_2} $ $ E_1 = E_2 $ $ \therefore {hc \over \lambda_1} = {nhc \over \lambda_2} $ $ \therefore n = {\lambda_1 \over \lambda_2} = {8000 \over 4000} = 2 $

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Output power of He-Ne LASER of low energy is 1.00 mW. Wavelength of the ligth is 632.8 nm. What will be the number of photons emitted per second from this LASER ?

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Explanation

$ = 1.0 mW = 10^{-3} W $ $ \lambda = 632.8 nm = 632 \times 10^{-9} m $ $\lambda = 632.8 nm = 632 \times 10^{-9} m $ $ P = { nhc \over \lambda } $ $ \therefore n = { p \lambda \over hc } $ $ = { 10^{-3} \times 6.3328 \times 10^{-7} \over 6.625 \times 10^{-34} \times 3 \times 10^ {3}} $ $ ={ 6.328 \times 10^{-20} \over 19.875 \times 10^{26} } $ $=0.318 \times 10^{16} $ $\therefore n = 318 \times 10^{15} s^{-1} $

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A star which can be seen withnaked eye from Earthhas intensity $ 1.6 \times 10^{-9} Wm^{-2} $ on Earth. If the corresponding wavelength is 560 nm, and the diameter of the human eye is$ 2.5 \times 10^{-3} m $ , the number of photons entering in our in 1 s is..............

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Explanation

$ 1 = 1.6 \times 10^ {-19} w/m^2 $ $ \lambda = 560 nm = 5.6 \times 10^ {-7} m $ $ r = 2.5 \times 10^{-3} m $ $ t = ls , n = ? $ $ 1 = { E \over At} = { P \over A } $ $ \therefore P = 1A ^\circ = 1( \pi r^2 ) = 1.6 \times 10^{-9} \times 3.14 \times 6.25 \times 10^ {-6} $ $ 31.4 \times 10^{-15} W $ $ \therefore = P = { nhc \over \lambda }$ $ \therefore = { p \lambda \over hc } = { 31.4 \times 10^ { -15 } \times 5.6 \times 10^{-7} \over 6.62 \times 10^{-34} \times 3 \times 10^3 }$ $ \therefore n = 8.85 \times 10^4 $

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What should be the ratio of de-Broglie wavelength of an atom of nitrogen gas at 300 K and 1000 K. Mass of nitrogen atom is $4.7 \times 10^{-26}$ kg and it is at 1 atm pressure Consider it as an ideal gas

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Explanation

$ T _1 = 300 K $ $ T_2 = 1000 K $ $ m = 4.7 \times 10^{-26} kg $ $ P = 1 atm $ $ { 1 \over 2 } mv ^2 = { 3 \over 2 } KT $ $ \therefore m^2 v^2 = p^2 = 3 mKT$ $ \therefore p = \sqrt {3mKT} $ $ \lambda = { h \over p } $ $ \therefore \lambda = { h \over \sqrt { 3mKT} }$ $ \therefore \lambda \alpha { 1 \over \sqrt T} $ $ \therefore { \lambda _1 \over \lambda_2 } = { \sqrt { T_2 \over T_1 }} = \sqrt { 1000 \over 300 } = \sqrt {10 \over 3 } $ $ \therefore { \lambda _1 \over \lambda_2 } = 1.826$

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