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

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

$at \,t = 0 \; \;\; N =No$ $ 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 $

Thomson मॉडल

$ {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$ $=0.0305 \times 931.48 = 28.4 MeV$

$ {1 \over \lambda } = R2^2[{1\over n_2^2} -{1\over n_1^2}] \leftarrow for \, any \,atom $

$ { 1 \over \lambda_1 (L) } = R(1)^2[{1\over 1^2} - {1 \over 2^2}]=RZ^2[{1\over 4} - { 1 \over 16 } ] {3RZ^2 \over 4}$

$ { 1 \over \lambda_1 (L) } = {3R \over 4} \Rightarrow \lambda_1(L) = { 4 \over 3R} $

$ { 1 \over \lambda_2 (B) } = R2^2[{1\over 2^2} - {1 \over 4^2}]=RZ^2[{1\over 4} - { 1 \over 16 } ] = {3RZ^2 \over 16}$

$ \Rightarrow \lambda _2 (B) = { 16 \over 3RZ^2} $ $ { \lambda_1 (L) \over \lambda _2 (B) }$ $ 1 = { 2^2 \over 4} \Rightarrow z^2= 4 Z=2$

संवेग संरक्षण के अनुसार फोटॉन का संवेग = प्रतिक्षेप परमाणु का संवेग। ${h \over \lambda} = m\vartheta $ $\therefore \vartheta = { h\over m \lambda}$

$$\begin{gathered} \overrightarrow{F}=-k\frac{e^2}{r^2}\hat{r} \left(Vector form\right) \\ =-k.\frac{e^2}{r^3}\overrightarrow{r} \end{gathered}$$    $\left(\because \hat{r}=\frac{{\displaystyle \overrightarrow{r}}}{{\displaystyle r}}\right)$  

$\frac{k{e}^2}{r^2}=\frac{m{v}^2}{r} \Rightarrow \frac{1}{2} m{v}^2=\frac{k{e}^2}{2r}$