Physics MCQs for NEET — Practice Questions with Answers

$A_1 = A_0 e^{-t_1/t}$ $A=A_0^ {e -\lambda t } = A = A_0e^{-t/t}$ $A_0 = {A_1 \over e-t_1/T} \Rightarrow A_0 = {A_1 e^{t_1 /T}} $ $ Now , A_2 = A_0e^{-t_2/T}$ $ =A_1 e^{t_1/T} e^{-t_2/T}$ $ A_2 = A_1 e(t_1 - t_2)/ T$

$_{92}U^{238} \rightarrow x \rightarrow _zY^A$ $_{92}U^{238} \rightarrow _{90}Th^{234} \rightarrow _{91}Y^{234}$

$_{92}U^{238} \rightarrow _{76}X^{206} \rightarrow _{82}Y^{206}$ $ \therefore _{82}Pb^{206}\; _{or}Pb^{206}$

Energy released per uranium atom = 200 MeV $= 200\times 10^6 ev = 200 \times 10^6 \times 1.6 \times 10 ^{-19}= 320\times10^{-13} \gamma $ $For 320 \times 10^{-13} 2 \rightarrow 1 atom$ $32 J \rightarrow \lambda {3.2 \over 320 \times 10^{-13}} =10^{11}atoms$

The energyleVel of hydrogen atom corresponding to 10.2 eV is n= 2, n = 1 correponding engery for the level of helium atom $ E = {-13.6 Z^2 \over n^2}$ $ \triangle E = E_2^1 - E_1^1 $ $ \triangle E = {13.6 \times 4 \over 4} - { (-13.6 \times 4) \over 1 } = -13.6 + 54.4 = 40.8 eV$

For any state K.E = $ {1 \over 8 \pi E_0} {Ze^2 \over Cn} and cangatoms 1$ $P.E = -{1 \over 4 \pi \epsilon_0 }{Ze^2 \over rn } (total energy in the state )$ E = K.E + P.E $E = -{ 1 \over {8 \pi \epsilon _0 }} {Ze^2 \over ru} = -K.E $

For wave length of second line of Balmer series $ {1\over \lambda_{2(B)}} = R [ {1\over 2^2} - {1\over 4^2}] = R [{1\over 4} - {1\over 16}] = {3R \over 16}$ $ \Rightarrow \lambda2(B) = {16\over 3R}$ For wave length of First line of Luman series $ {1\over \lambda_{2(L)}} = R [ {1\over 1^2} - {1\over 2^2}] = R [{1\over 1} - {1\over 4}] = {3R \over 4}$ $ \lambda_1(L) = {4\over 3R}$ $ {\lambda_2(B) \over \lambda_1(L)} = {16 \over 3R} \times {3R \over 4 } = 4 $ $ \lambda_1(C) = {\lambda_2(B)\over 4} = {486.4 \over 4 } = 121.6 mm$

$ r \,\alpha {n^2 \over z } $ For hydrogen atom Z = 1 $ r \,\alpha \, n^2 $ $ {r_1 \over r_2 } = {(1 \over 4 )}$ $ r_2 = 4r_1 = 4 \times 0.53 A = 2.12 A $

According comservation of momentum momentum of photon = momentum of rocil atom. ${h \over \lambda} = m\vartheta $ $\therefore \vartheta = { h\over m \lambda}$

average speed is the ratio of total distance travelled to the total time taken .since initial and final position is the same average velocity is zero