Physics MCQs for NEET — Practice Questions with Answers

$ \phi = hf_0 = { hc \over \lambda_0 e } eV $ $ \therefore \phi \alpha { 1 \over \lambda_0}$ $ \therefore { \phi_T \over \phi _N } = { (\lambda_0) _N \over (\lambda_0)_T } =5460 \times {2.3 \over 4.5} $ =2791 A $ c = f_0 \lambda_0 $ $ \therefore f_0 = { c \over \lambda_0 } = { 3 \times 10^{8} \over 2791 \times 10^{-10} } $ $ = 1.075 \times 10^{15} Hz$ $ f_0 = 1.075 \times 10^{15} Hz$

$ P = { h \over \lambda } $ $ \therefore P \alpha { 1 \over \lambda }$ $ \therefore { P_1 \over P_2 } = { \lambda_1 \over \lambda_2 } = { 8000 \over 4000} = { 2 \over 1 } $ $ \therefore {P_1 \over P_2} = 2:1 $

$ { 1 \over 2 } mv^2 max = V_0 e = hf - hf_0 $ $ \therefore V_0 = { hf - hf_0 \over e } = { (6.2 -4.2 ) \times 1.6 \times 10^{-19} \over 1.6 \times 10^{-19} }$ $ \therefore V_0 = 2V $

According to the photoelectric effect, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. Higher intensity means more photons hitting the surface, thus more photoelectrons are emitted.

E = hf $ \therefore f = { E \over h } = { 66 \times 1.6 \times 10^{-19} \over 6.6 \times 10^{-34}} $ $ = 16 \times 10^{15} Hz $

The kinetic energy (K.E.) of a proton when it is accelerated through a potential difference (p.d.) of 1 V is calculated using the equation K.E. = qV, where q is the charge of the proton. Since the charge of a proton is approximately 1e (1 electron volt), the kinetic energy will be 1 eV.

Polarisation is a phenomenon that can be explained by the wave theory of light, not by the quantum theory of light. The quantum theory explains phenomena like the photoelectric effect, black body radiation, and the Compton effect.

$ m = { m_0 \over \sqrt {1 -{v^2 \over c^2 }} } = 2m_0 $ $ \therefore \sqrt {1 -{v^2 \over c^2 }}= { 1 \over 2 } $ $ {1 -{v^2 \over c^2 }} = {1 \over 4 } $ $ \therefore {v^2 \over c^2 } = { 3 \over 4 } $ $ \therefore { v \over c }= { \sqrt 3 \over 2 }$ $ \therefore v = { \sqrt3 \over 2 } e $

$ \lambda = { h \over p } $ $ \therefore \lambda \alpha {1 \over p} $ $ \therefore \lambda \alpha p^{-1} $ $ d \lambda \alpha -{p ^ {-2}}dp $

$ d \lambda \alpha -{ 1 \over p ^ {-2}}dp $ $ \therefore { dp \over \lambda } \times 100 = -{ dp \over p^2} \times p \times 100 $ $\therefore { dp \over \lambda } \times 100 = -1\%$ $ decreases by \%$

$ \therefore = { h \over \sqrt { 2Vem}} $ $ v = { h^2 \over 2me \lambda ^2 } $ $ = { (6.62 \times 10^{-34})^2 \over 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times (0.4 \times 10^{-1} )^2 }$ = 940.5 =941 V