NEET Practice Questions (MCQs) with Answers & Solutions

To find the time required for the photoelectric effect to start, we need to calculate the energy absorbed by the electrons and then compare it with the work function of the metal. The intensity of incident light is given as $10^{-5} ext{ Wm}^{-2}$, and the area of the metal is $2 ext{ cm}^2$ or $2 imes 10^{-4} ext{ m}^2$.

$ t -10 s ,N = 11 \times 10^{11}, \lambda = 10A , A = 0.01 m^2 $ $ 1 = { E \over A.t} $ $ E = { NHc \over \lambda } $ $ \therefore I = { NHc \over \lambda At } $ $ \therefore I = { 11 \times 10^ {11} \times 6.625 \times 10^ {-34} \times 3 \times 10^{8} \over 10 \times 10^ {-10 } \times 10^{-2} \times 10 } $ $ = 218.6 \times 10^{-5} $ $ = 2.186 \times 10^{-3} w/m^2 $

$ \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 $