NEET Practice Questions (MCQs) with Answers & Solutions

$ { 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 $

final friquency $ { 1.5 \over 2 } f_0 = 0.75 f_0 < f_0 $ therefore photoelectry current is zero

$ m = { m_0 \over \sqrt {1 -{v^2 \over c ^2 }} } = { m_o \over \sqrt { 1 - { 0.64 \times 9 \times 10^{16} \over 9 \times 10^{16} }}}$ $ = { m_0 \over \sqrt 0.36 } = { m_0 \over 0.6 } = { 10m_0 \over 6 } = { 5m_0 \over 3 } $

The kinetic energy (K.E) gained by a charged particle when accelerated through a potential difference V is given by $K.E = qV$. Here the charge of the particle is double that of an electron (2e) and the potential difference is 5V. So, the K.E = $2e \times 5V = 10$ eV.

$ { 1 \over 2 } mv^2 max = { hc \over \lambda } - \phi $ $ = ( {6.62 \times 10^{-34} \times 3.0 \times 10^8 \over 3.0 \times 10^{-7} \times 1.6 \times 10^{-19}} -1) eV $ $ = 3.1375 eV = 5.02 \times 10 h -19 J $ $ \therefore V_{max} = \sqrt { 2 \times 5.0 \times 2 \times 10^{-19} \over 9.1 \times 10^{-31 } }$ $ ( \therefore m = 9.1 \times 10^{-31} kg )$ $ = 1.05 \times 10^6 = 1 \times 10^6 ms^{-1} J$

$ \phi = hf _0 $ $ = { hc \over \lambda_0 } J = { hc \over \lambda_0 e } eV $ $ \therefore \lambda_0 \alpha {1 \over 0 } $ $ \therefore {\varphi_0( tungsten) \over \varphi _0 (sodium) }= { \varphi (tungsten) \over \varphi (sodium) } ={ 2.3 \over 4.5} $ $ \therefore \lambda_0 tungsten = { 2.3 \over 4.5 } \times 5460 A ^ \circ $ $ = 2790.6 = 2791 A ^ \circ $

In the Davisson and Germer experiment, the filament of the electron gun is coated with barium oxide (BaO). This coating helps in the efficient emission of electrons when heated.