NEET Practice Questions (MCQs) with Answers & Solutions

$ \lambda_1 = 3500 A ^\circ , \lambda _2 =290 nm $ $ h = 6.625 \times 10^ { -24} J.s $ $ { 1 \over 2 } mv^2 max = { hc \lambda } - \phi = eV_0 $ $ \therefore V_01 e = { hc \over \lambda _1 } -\phi $ $ \therefore V_02e = { hc \over \lambda_1} -\phi $ $ \therefore V_02 e = { hc \over \lambda_2} - \phi $ $ (V_02 -V_01) e = hc ( {1\over \lambda_2 } - { 1 \over \lambda_1} )$ $ \therefore V_02 -V_01 = { hc \over e } [ { \lambda_1 - \lambda_2 \over \lambda_1 \lambda_2} ] = 12.42 = [ { 0.6 \over 3.5 \times 2.9 } ]$ $ = 0.7342$ $ = 73.42 \times 10^{-2} V $

$ \lambda = 6625 A^\circ $ $ = 6.625 \times 10^ { -7} m $ $ c = 3 \times 10^8 m/s $ $ E = nhf $ $ E = { nhc \over \lambda } $ $ n = { E \lambda \over hc } = { 3 \times 6.625 \times 10^{-7} \over 6.625 \times 10^{-34} \times 3 \times 10^ 8 }$ $ \therefore n = 10^{19 } $

$ eV_0 = { hc \over \lambda } - \phi $ $ \therefore V_0 = { hc \over \lambda c } = { \phi \over e } $ $ \therefore V_0 = 1031.4V $

$ \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 $ $= 9.55 \times 10^ 9 eV = 9.55 \times 10^ { 3} MeV$

To find the de Broglie wavelength of a proton after falling freely under gravity for 10 seconds, we first calculate the velocity (v) using the equation of motion: $$v = u + gt$$, where initial velocity (u) is 0, g is 10 m/s², and t is 10 s. Thus, $$v = 0 + (10 m/s² imes 10 s) = 100 m/s$$. The momentum (p) of the proton is given by: $$p = mv$$, where m is the mass of the proton, $$p = 1.6 imes 10^{-27} kg imes 100 m/s = 1.6 imes 10^{-25} kg imes m/s$$. The de Broglie wavelength λ is: $$ ext{λ} = rac{h}{p} = rac{6.625 imes 10^{-34} Js}{1.6 imes 10^{-25} kg imes m/s} ext{λ} = 41.40625 imes 10^{-10} m = 41.40625 Å$$. This is closest to 39.6 Å, so the correct option is $$39.6 Å$$.

$ 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 $

$E = K + U $ $ E = { p ^ 2 \over 2m } - eV_{(x)} $ $ \therefore p = ( 2mE + eV_{(x)} )^{1/2} $ $ \therefore \lambda = { h \over p } = { h \over [ 2m ( E + eV_{(x)} )]^{1/2} }$ $ \mu = { \lambda_0 \over \lambda } = { h \over \sqrt { 2mE} } \times { [2m ( E + eV_{(x)})]^{1/2} \over h }$ $ \mu = [ { E + eV_{(x)} \over E} ] ^ { 1 /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 $