Atoms MCQs for NEET — Physics Questions with Answers

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 $ $\therefore \vartheta = { h\over m \lambda}$

$ {N \over N0} = ( {1\over 2} ) ^ {t /T1/2} = ({ 1 \over2}) {72000 \over 24000} = {1 \over 8} $

$at \,t = 0 \; \;\; N =No$ $ t = T_ {1/2} \times 1 \;\;\; N = No/2 \;\;\; \therefore {N \over No} = {1 \over 16}$ $ t = T_ {1/2} \times 2 ;;; N = No/4 ;;; {N\over No} \times 100 = { 1 \over 32 } \times 100$ $ t = T_ {1/2} \times 3 \;\;\; N = No/8 \;\;\; =3.125 $ $ t = T_ {1/2} \times 4 \;\;\; N = No/16 \;\;\; \approx 3 \%$ $ t = T_ {1/2} \times 5 \;\;\; N = No/32 $

$ wave number = {1 \over \lambda} = R [{1 \over 2^2} - {1 \over 3^2}] = R[{1\over 4} - {1 \over 9}] = { 5R \over 36 }$

$mass \, defect \, \triangle m = ( 2mp +2mn) -MHe$ $ \triangle m = ( 2 \times 1.0087 -2 \times 1.0073) - 4.0015 $ = 4.032-4.0015 $ \triangle m = 0.030 amu$ $ E = \triangle m \times 931.48 mev$ $=0.0305 \times 931.48 = 28.4 MeV$