Dual Nature of Matter and Radiation MCQs for NEET — Physics Questions with Answers

$ 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 $

$ \lambda = { h\over p} = { h \over mv} $ $ m = { m_0 \over \sqrt {1 -{v^2 \over c^2}} }$ $ \lambda = { h ( { \sqrt {1 - { v^2 \over c^2 }}}) \over m_0 v}$ $ v ={3 \over 4 } $ $ \lambda = {\lambda \sqrt {1 -{c^2 \over 16 c^2}} \over m_0 c/4}$ $ \lambda = {h \sqrt {16c^2 -c^2 \over 16 c^2} \over m_0 c/4}$ $ = { \sqrt { 15/16 } h \over m_0 c/4} = { 4 \times 0.968 h \over m_0 c } $ $ \therefore \lambda = { 3.87 h \over m_0c }$

$ \phi = 2.5 eV = 2.5 \times 1.6 \times 10^{-19} J$ $ \lambda = 3000 A = 3 \times 10^{-7} m $ $ h = 6.62 \times 10^{-34} J.s$ $ c = 3 \times 10^8 J.s $ $ { 1 \over 2 } mv^2 _{max} = eV_0 = { hc \over \lambda } - \phi $ $ \therefore V_0 = { hc \over \lambda e } - { \phi \over e } $

$ \phi = hf_0 = { hc \over \lambda_0 e } eV $ $ \therefore \phi \alpha { 1 \over \lambda_0}$ $ \therefore { \phi_T \over \phi _N } = { (\lambda_0) _N \over (\lambda_0)_T } =5460 \times {2.3 \over 4.5} $ =2791 A $ c = f_0 \lambda_0 $ $ \therefore f_0 = { c \over \lambda_0 } = { 3 \times 10^{8} \over 2791 \times 10^{-10} } $ $ = 1.075 \times 10^{15} Hz$ $ f_0 = 1.075 \times 10^{15} Hz$

$ P = { h \over \lambda } $ $ \therefore P \alpha { 1 \over \lambda }$ $ \therefore { P_1 \over P_2 } = { \lambda_1 \over \lambda_2 } = { 8000 \over 4000} = { 2 \over 1 } $ $ \therefore {P_1 \over P_2} = 2:1 $

$ { 1 \over 2 } mv^2 max = V_0 e = hf - hf_0 $ $ \therefore V_0 = { hf - hf_0 \over e } = { (6.2 -4.2 ) \times 1.6 \times 10^{-19} \over 1.6 \times 10^{-19} }$ $ \therefore V_0 = 2V $