The 2s-orbital of H-atom has a radial node at 2a0 because $${\psi _{2s}}$$ is proportional to
A. $$\left( {\frac{1}{2} + \frac{r}{{{a_0}}}} \right)$$
B. $$\left( {2 + \frac{r}{{{a_0}}}} \right)$$
C. $$\left( {2 - \frac{r}{{{a_0}}}} \right)$$
D. $$\left( {2 - \frac{r}{{2{a_0}}}} \right)$$
Answer: Option C
A. 99.7 × 10-12 m
B. 199.4 × 10-12 m
C. 199.4 × 10-18 m
D. 99 × 10-6 m
A. $$\exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right)$$
B. $${\left[ {1 - \exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right)} \right]^{ - 1}}$$
C. $$\exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right){\left[ {1 - \exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right)} \right]^{ - 1}}$$
D. $$\exp \left( { - \frac{{h\nu }}{{2{K_B}T}}} \right){\left[ {1 - \exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right)} \right]^{ - 1}}$$
A. 4 × 104 (nm)2
B. 10$$\sqrt 2 $$ (nm)1/2
C. $$\sqrt 2 $$ /10 (nm)-1/2
D. 0.1 (nm)-1/2
A. $$\Delta \varepsilon _n^{\left( 1 \right)} = \gamma $$
B. $$\Delta \varepsilon _n^{\left( 1 \right)} = {\gamma ^2}$$
C. $$\Delta \varepsilon _n^{\left( 1 \right)} = {\gamma ^{ - 1}}$$
D. $$\Delta \varepsilon _n^{\left( 1 \right)} = 0$$

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