Dual Nature of Matter and Radiation NEET प्रश्न हिंदी में — भौतिक विज्ञान (Physics) MCQ उत्तर सहित

$ \triangle x = 2 r = 2 \times 10^ {-15 } m $ $ \triangle . \triangle p = { h \over 2 \pi } $ $ \therefore \triangle p = { h \over 2 \pi \triangle x } = 0.5274 \times 10^ { -19} $ $ E = { p^2 \over 2m } P = \triangle p $ $= 9.55 \times 10^ 9 eV = 9.55 \times 10^ { 3} MeV$

$ 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 $

$ {1 \over 2 } mv^2 max = eV_0 = { hc \over \theta } - \phi $ $ = { 6.625 \times 10^ {-34} \times 3 \times 10^8 \over 2 \times 10^{-7} }$ $ = 4.5 \times 1.6 \times 10^ {-19} $ $ \therefore V_0 = { 9.94 \times 10^{-19} \over 1.6 \times 10^{-19} } = { 4.5 \times 1.6 \times 10^{-19 } \over 1.6 \times {-19 } } $ $ = 6.21 -4.5 $ $ \therefore V_0 = 1.71 V $ $ { 1\over 2} mv^2 max = eV_0 = 1.71 \times 1.6 \times 10^{-19} $ $ \therefore v^2 max = { 2 \times 2.74 \times 10^{-19} \over m } $ $ \therefore V_{max} = \sqrt { 5.48 \times 10^{-19} \over 9.11 \times 10^{-31} $ $ \therefore V_{max} = 7.75 \times 10^5 m/s $

$ V = 50 KV = 50 \times 10^3 V $ $ \lambda = { h \over \sqrt { 2meV}} = { 6.62 \times 10^{-34} \over \sqrt { 2 \times 9.1 \times 10^{-31} \times 50 \times 10^{3} \times 1.6 \times 10^{-19}}} $ $ = {6.62 \times 10^{-34} \over 1.207 \times 10^{-22}} $ $ = 5.485 \times 10^{-12} m $

$ 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$

$ t -10 s ,N = 11 \times 10^{11}, \lambda = 10A , A = 0.01 m^2 $ $ 1 = { E \over A.t} $ $ E = { NHc \over \lambda } $ $ \therefore I = { NHc \over \lambda At } $ $ \therefore I = { 11 \times 10^ {11} \times 6.625 \times 10^ {-34} \times 3 \times 10^{8} \over 10 \times 10^ {-10 } \times 10^{-2} \times 10 } $ $ = 218.6 \times 10^{-5} $ $ = 2.186 \times 10^{-3} w/m^2 $

$ E_p = { hc \over \lambda} $ $ E_p = 19.87 \times 10^{-16} J $ $ E_0 = { p^2 \over 2m } = { h^2 \over \lambda^2 (2m) } $ $ \therefore E_0 = 2.41 \times 10^{-17 } J $ $ \therefore { E_p \over E_0 } = { 19.87 \times 10^{-16} \over 2.41 \times 10^ {-17 } } $ $ \therefore { E_p \over E_0} = 82.4 $