Physics MCQs for NEET — Practice Questions with Answers

$ \lambda = { h \over mv } $ $ \therefore \lambda_\infty = { h \over m_\infty v_\infty }$ $\lambda_d = { h \over m_d v_d} $ $ \therefore { \lambda_\infty \over \lambda_d } = { m_d v_d \over m_\infty v_\infty} = { 2 \times 2v \over 4 \times v } $ $ \therefore {\lambda_\infty \over \lambda_d } = 1 $

According to Heisenberg's uncertainty principle, the uncertainty in position ( ext{∆x}) and momentum ( ext{∆p}) is given by the relation: $$ ext{∆x ∆p} hicksim rac{h}{2 ext{π}} $$

According to the photoelectric effect equation, the maximum kinetic energy of emitted photoelectrons is given by: $$ K.E. = hf - ext{Work Function} $$ Given that the frequency of light is $f_1$ and $f_2$ with $f_1 > f_2$, and the ratio of maximum kinetic energy is 1:K, we can write two equations: $$ K.E._1 = hf_1 - ext{Work Function} $$ $$ K.E._2 = hf_2 - ext{Work Function} $$ Given the ratio of kinetic energies: $$ rac{K.E._1}{K.E._2} = K $$ Substitute the kinetic energy equations: $$ rac{hf_1 - ext{Work Function}}{hf_2 - ext{Work Function}} = K $$ Let the work function be $hf_0$, where $f_0$ is the threshold frequency: $$ rac{hf_1 - hf_0}{hf_2 - hf_0} = K $$ Simplifying this equation, we get: $$ rac{f_1 - f_0}{f_2 - f_0} = K $$ Solving for $f_0$ (threshold frequency): $$ K(f_2 - f_0) = f_1 - f_0 $$ $$ Kf_2 - Kf_0 = f_1 - f_0 $$ $$ f_0(K - 1) = Kf_2 - f_1 $$ $$ f_0 = rac{Kf_2 - f_1}{K - 1} $$

$ 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 } $