Physics MCQs for NEET — Practice Questions with Answers

$ \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$

According to Einstein's photoelectric equation, the graph of the kinetic energy of emitted photoelectrons versus the frequency of incident radiation is linear, and the slope of this graph is given by Planck's constant $h$. This slope is the same for all metals and is independent of the intensity of the radiation. Therefore, the correct answer is that the slope is the same for all metals and is free from the intensity of radiation.

I is inversely proportional to 1/r^2

According to de-Broglie's hypothesis, the wavelength λ is inversely proportional to the momentum p of the particle: $$ ext{λ} = rac{h}{p}$$ where h is Planck's constant. For a free electron, the kinetic energy (K.E.) is given by: $$ ext{K.E.} = rac{p^2}{2m}$$ where m is the mass of the electron. If the kinetic energy is doubled, $$ ext{K.E.}' = 2 imes rac{p^2}{2m} ightarrow ext{p}' = ext{p} imes rac{1}{ ext{√}2}$$. Hence, the new de-Broglie wavelength λ' will be: $$ ext{λ}' = rac{h}{ ext{p}'} = rac{h}{ ext{p} imes rac{1}{ ext{√}2}} = ext{λ} imes ext{√}2$$. Thus, the change in de-Broglie wavelength will be inversely proportional to √2, making the correct answer $$rac{1}{ ext{√}2}$$.

The energy of the incident radiation can be determined using the photoelectric effect equation: $$ ext{Energy of incident radiation} = ext{Threshold energy} + ext{Kinetic energy of ejected electrons}$$. Given the threshold energy is 6.2 eV and the stopping potential is 5V, the total energy of the incident radiation will be: $$6.2 ext{ eV} + 5 ext{ eV} = 11.2 ext{ eV}$$. The corresponding wavelength for this energy falls within the ultraviolet (UV) region of the electromagnetic spectrum.

$ T_1 = 10 +273 = 283 K $ $ T_2 = 40 +273 = 313 K $ $ \lambda _1 = 0.4 A$ $ \lambda = { h \over \sqrt {2mE} }$ $ E = {3 /2 } KT $ $ \lambda = { h \over \sqrt { 3mKT}}$ $ \therefore \lambda \alpha { 1 \over \sqrt T} $ $ \therefore { \lambda_1 \over \lambda_2} = \sqrt { T_1 \over T_2}$ $ \therefore {\lambda_2 \over 0.4 \times 10^{-10}} = \sqrt {283 \over313}$ $ \therefore {\lambda_2 \over 0.4 \times 10^{-10} }= 0.951 $ $ \therefore \lambda_2 = 0.951 \times 0.4 \times 10^{-10}$ $ \therefore \lambda _2 = 0.38 A $ $ = \lambda_2 - \lambda_1 $ $ = 0.38 A ^\circ - 0.4 A ^\circ = - 0.02 A ^\circ $ $ = 2 \times 10^ {-2} A ^\circ decreases $

$ \lambda_1 = 4000 A^ \circ = 4 \times 10^{-7}$ $ \lambda _2 = 3600 A^ \circ = 3.6 \times 10^{-7} m $ $V_0 e = { hc \over \lambda } - \phi .....(1)$ $ \therefore V_{01}e = { hc \over \lambda_2} -\phi.....(2)$ $ V_{02}e - V_{01}e = hc [{1 \over \lambda_2} - {1 \over \lambda_2} ]$ $ \therefore V_{02} -V_{01} = { hc \over e} [{ 1 \over \lambda_2 } - {1 \over \lambda _1} ] $ $ = { 6.62 \times 10^{-34} \times 3 \times 10^8 \over 1.6 \times 10^{-19}}[{10^7 \over 3.6} - {10^7 \over 4 } ]$ $ = {6.62 \times 3 \over 1.6 } [{4-3.6 \over 4 \times 3.6 }] $ $ \therefore V_{02} -V_{01} = 0.345 V$

$\lambda_1 = 4000 A^\circ = 4 \times 10^ -7 m $ $ \lambda_2 = 3000 A^\circ = 3 \times 10 ^ {-7} m$ $ \triangle = ? $ $ E = { hc \over \lambda } -\phi $ $ E_1 = { hc \over \lambda_1} -\phi ...(1) $ $ E_2 = { hc \over \lambda_2} - \phi...(2) $ $ \therefore E_2 -E_1 = hc ({1 \over \lambda_2 } - { 1 \over \lambda_1} )$ $ = 6.625 \times 10^{-34} \times 3 \times 10^{8} [{10^7 \over 3} - {10^7 \over 4}] = 19.956 \times 10^{-19} [{1 \over 12}]$ $ \therefore \triangle E = 1.04 eV$