भौतिक विज्ञान (Physics) NEET प्रश्न हिंदी में — उत्तर सहित MCQ अभ्यास

माध्य जीवन (T) में विश्राम द्रव्यमान = $No/ e$ $ T = { 1 \over \lambda } \Rightarrow \lambda = {1 \over 5 } min ^{-1}$ $ N = N0e^{- \lambda t}$ $ N0/2 = N0e^{-1/5 t}$ $m2 = { 1\over 5}t $ $ {1 \over 2 } = e^{-1/5} t $ $ t = 5ln2$ $ 2^{-1} = e^{-1/5} t $ $ t = 5 log e^2$ $ 2^1 = e {1 \over 5 } t $

${dN \over df} = df \Rightarrow I = dN$ $ I^1 = \lambda N_1 $ $ I_2 = \lambda N_2 $ $ {I_1 \over N_1} = \lambda $ $ {I_1 \over N_2} = \lambda $ $\therefore {I_1 \over N_1} = { f_2 \over N_2} \Rightarrow I_2 = { I_2 \over N_1} \times N_2 = { 3 \times 10^{10} \times 2 \times 10^{15} \over 2 \times 10_20} = 3 \times 10^5 $

Thomson मॉडल

$\triangle E= E_4 - E_1 = {-13.6 \over 4} - ({-13.6 \over 1})= 0.85 +13.6 = 12.75 ev$

$ R = Ro(a)^{1/3} = 2R = Ro(A){1/3}$ $ \therefore { 1 \over 2 } = ({ 9 \over A }) ^ { 1/3} = { 1 \over 8 } = { 9 \over A} \Rightarrow \therefore A =72 $

$ {1 \over \lambda} R [{ 1 \over 1 ^2 } - { 1 \over n^2 } ] $ $ {1 \over \lambda } = R [ 1- { 1 \over n^2 } ]$ $ { 1 \over \lambda } = R - { R/ n^2}$ $ { R \over n^2 }= R -{1/ \lambda } = { \lambda R -1 \over \lambda }$ $ { R \over n^2 } = { \lambda R -1 \over \lambda}$ $ n^2 = { \lambda R \over \lambda R -1 }$ $ n = \sqrt {\lambda R \over \lambda R -1}$

$mass \, defect \, \triangle m = ( 2mp +2mn) -MHe$ $ \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$ $=0.0305 \times 931.48 = 28.4 MeV$

Vsq=i3×1+rI3

जहाँ Vsq=26/

अतः, r = 25Ω