Physics MCQs for NEET — Practice Questions with Answers

$ \lambda_p = \lambda_\alpha $ $ \therefore { h \over m_p v_\alpha } = { h \over m_\alpha v_\alpha } $ $ \therefore { v_p \over v_\alpha} = { m_\alpha \over m_p} $ $ m_\alpha = 4m_p $ $ { v_p \over v_\alpha } = { 4m_p \over m_p} = 4 $ $ \therefore {v_p \over v_\alpha} = 4:1 $

$ \lambda = { h \over mv } $ $ m = { m_0 \over \sqrt {1 - { v^2 \over c^2}} }$ $ \lambda = { h \times \sqrt { 1 - { v^2 \over c^2 } \over m_0 v }} $ $ v =c , \lambda = { h \times \sqrt { 1 - { c^2 \over c^2 } \over m_0 v }} $ $ \therefore \lambda = 0 $

$ ( {1 \over 2 mv^2 _{max} }) _1 = (hf)_1 - \phi = 1 - 0.5 $ $ ( {1 \over 2 mv^2_ {max} }) _2 = (hf)_2 - \phi = 2.5 - 0.5 $ $ \therefore {(v^2 _{max} )_1 \over (v^2 _ {max }) _2 } = { 0.5 \over 2 } = { 1 \over 4 } $ $ \therefore {(v^2 _{max} )_1 \over (v^2 _ {max }) _2 } = { 1 \over 2 } = 1:2 $

$ m_p = 1 q_p = e $ $ \lambda = { h \over \sqrt {2meV}} $ $ \therefore \lambda \alpha { 1 \over \sqrt {mq} } $ $ \therefore { \lambda_p \over \lambda_\alpha } = \sqrt { m_\infty q _ \infty \over m_p q _p }$ $ = \sqrt { 4 \times 2e \over 1 \times e } $ $ = \sqrt 8 $ $ = 2 \sqrt 2 $ $ \therefore { \lambda _p \over \lambda_\alpha } = 2 \sqrt {2} : 1 $

$ \triangle x .\triangle p = { h \over 2 \pi } = h $ $ \triangle x = \lambda $ $ \triangle = {h \over \lambda} $

$ \lambda = { h \over \sqrt {2mE} } = { 6.6 \times 10^{-34} \over \sqrt { 2 \times 9.1 \times 10^{-31} \times 10^4 \times 1.6 \times 10^{-19} }}$ $ = 0.12 \times 10^{-10} m $ $ \lambda = 0.12 A ^\circ $

$ { K.E of practical \over K.E of photon} = { {1 \over 2 } mv^2 \over hf } = { {1 \over 2 } mv.v \over {hc \over \lambda} } = { {1 \over 2 } Pv \lambda \over hc }$ $ = { {1 \over 2 } {h \over \lambda } \lambda v \over hc } = { v \over 2c} ={ 2.25 \times 10^8 \over 2 \times 3 \times 10^8 } = {3 \over 8 } $

$ { 1 \over 2 } mv_m^2 = { hc \over \lambda} - \phi $ $ = 2 - \phi (eV) ....(1) $ $ {1 \over 2 } m 4 v_m^2 = { hc \over {3 \over 4 } \lambda } - \phi $ $ = { 4 \over 3 } \times 2 - \phi = {8 \over 3 } -\phi $ $ \therefore 4 ( { 1 \over 2 } mv_m^2 ) = { 8 \over 3 } - \phi $