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

In the Davisson and Germer experiment, the velocity of the electrons emitted from the electron gun can be increased by increasing the potential difference between the anode and the filament. A higher potential difference accelerates the electrons to higher velocities.

Here, surface is perfacet reflector momentum of incident radiation is E/C momentum of reflected rediation is - E/C change in momentum = - Momentum transtered to the surface = $2E \over C$

$ W = {1 \over 2} mv^2 = q V $ $ \therefore v = \sqrt {2qV/m} $ $ \therefore {V_A \over V_B} = \sqrt {q_A \over q_B} = \sqrt { q \over 4q} = 1/2 $ $ \therefore { V_A \over V_B } = 1:2 $

$ {1 \over 2 } mv^2_{max} = {hc \over \lambda } - \phi ....(1) $ $ \lambda' = \lambda - 0.25 \lambda = 0.75\lambda \,and \,v' = 2v $ $ \therefore {1 \over 2 } m ( 2v_{max})^2 = { hc \over 0.75 \lambda } -\phi.....(1)$ $4 ( { \lambda c \over \lambda } - \phi) = {4hc \over 3\lambda } - \phi $ $ \therefore {8hc \over 3\lambda} = 3 \phi $ $ \therefore \phi = { 8hc \over 9\lambda } $ $ { hc \over \lambda } = 2eV $ $ \phi = { 8 \over 9 } \times 2eV = 16/9 $ $ \phi = 1.8 eV$

$ p = { nhc \over \lambda } $ $ \therefore n = { p \lambda \over hc } = { 100 \times 540 \times 10^{-9} \over 6 \times 10^{-34} \times 3 \times 10^8 }$ $ = 3 \times 10^{20} $

$ {1 \over 2 } mv^2 = eV $ $ \therefore m^2 v^2 = p^2 =2meV$ $ \therefore p = {h \over \sqrt {2meV}}$ $ \therefore p \alpha { 1 \over \sqrt {me}} $ $ \therefore {\lambda_\infty \over \lambda _p } = \sqrt { m_p e_p \over m_\infty e_\infty} $ $ m_\infty = 4 mp ,e_\infty = 2 e_p $ $ \therefore { \lambda _\infty \over \lambda_p} = \sqrt{ m_p e _p \over 4m_p \times 2 e_p } = {1 \over \sqrt 8 } $ $ {\lambda_\infty \over \lambda _p } = { 1 \over 2\sqrt 2} $ $ \therefore {\lambda_\infty \over \lambda_0 } = { 1 \over 2\sqrt 2} $ $ \lambda _8 = {\lambda _0 \over 2\sqrt 2}$

$ \phi = 4eV = 4 \times 1.6 \times 10^{-19} J$ $ \therefore \phi = { hc \over \lambda_0} $ $ \therefore \lambda_0 = { hc \over \phi} $ $ ={6.62 \times 10^{-34} \times 3 \times 10^8 \over 4 \times 1.6 \times 10^{-19}}$ $ = 3.103 \times 10^ {-7} m$ $ =310.3 \times 10^ {-9} m$ $ = 310 m $

$ E = { 1/2} mv^2 max = hf - hf _0 = h (f-f_0) $ $ E_1 = h(f_1-f_0)$ $ E_2 = h (f_2 -f_0)$ $ f_1 = 5.5 \times 10^8 MHz = 5.5 \times 10^{14} Hz $ $ f_2 = 4.5 \times 10^ 8 MHz = 4.5 \times 10^{14} MHz $ $ {E_1 \over E_2} = { f_1 -f_0 \over f_2 - f_0} $ $ { E_1 \over E_2 } = { 1 \over 5 } $ $ \therefore { 1 \over 5 }= { 5.5 \times 10^{14} -f_0 \over 4.5 \times 10^{14} -f_0 } $
$ \therefore 4.5 \times 10^{14} - f_0 = 27.5 \times 10^{14} -5f_0$ $ \therefore 4f_0 = 23.0 \times 10^{14} $ $ \therefore f_0 = { 23 \times 10 ^ { 14} \over 4 }$ $ = 5.75 \times 10^14 Hz$ $ = 5.75 \times 10^8 Hz$

$ \lambda = { \lambda \over p } ...(1) $ $ \therefore \lambda + { 0.25 \over 100} \lambda = { h \over p-p_0} $ $ \lambda {100.25 \lambda \over 100 } = { p \over p-p_0} $ $ \therefore 100.25 p - 100.25 p_0 = 100 p $ $ \therefore 0.25 p =100.25 p_0 $ $ \therefore p = { 100.25 \over 0.25 } $ $ \therefore p = 401 p_0$