The energy required by electrons, present in the first Bohr orbit of hydrogen atom to be excited to second Bohr orbit is ____ Jtext mol^-1$J\text{ mol}^{-1}$. Given: R_H = 2.18 times 10^-11text ergs$R_{H} = 2.18 \times 10^{-11}\text{ ergs}$.

### Related Formula frac1lambda = RZ^2left(frac1n_1^2 - frac1n_2^2right)$$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ ### Core Logic Statement I is a factual description of how the hydrogen emission spectrum is obtained. Dissociation yields excited atoms that emit light at discrete frequencies. So Statement I is true. For Statement II: First line of Lyman series for H atom (Z=1, n_1=1, n_2=2$Z=1, n_1=1, n_2=2$): frac1lambda_1 = R(1)^2left(frac11^2 - frac12^2right) = Rleft(1 - frac14right) = frac3R4$$\frac{1}{\lambda_1} = R(1)^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$$ Second line of Balmer series for He^+$He^+$ (Z=2, n_1=2, n_2=4$Z=2, n_1=2, n_2=4$): frac1lambda_2 = R(2)^2left(frac12^2 - frac14^2right) = 4Rleft(frac14 - frac116right) = 4Rleft(frac316right) = frac3R4$$\frac{1}{\lambda_2} = R(2)^2\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4R\left(\frac{1}{4} - \frac{1}{16}\right) = 4R\left(\frac{3}{16}\right) = \frac{3R}{4}$$ Since frac1lambda_1 = frac1lambda_2$\frac{1}{\lambda_1} = \frac{1}{\lambda_2}$, their wavelengths (and therefore frequencies) are exactly the same. Statement II is true. ### Step 1: Final Conclusion Both statements are correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula Delta E = 13.6 Z^2 left(frac1n_1^2 - frac1n_2^2right) text eV$$\Delta E = 13.6 Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \text{ eV}$$ ### Core Logic Let's find the values of n_1$n_1$ and n_2$n_2$ for each transition: - (A) Paschen (1^st$1^{st}$ line): n_1 = 3, n_2 = 4$n_1 = 3, n_2 = 4$ - (B) Balmer (2^nd$2^{nd}$ line): n_1 = 2, n_2 = 4$n_1 = 2, n_2 = 4$ - (C) Paschen (3^rd$3^{rd}$ line): n_1 = 3, n_2 = 6$n_1 = 3, n_2 = 6$ - (D) Bracket (4^th$4^{th}$ line): n_1 = 4, n_2 = 8$n_1 = 4, n_2 = 8$ Calculating or comparing the energy values corresponding to these transitions yields the ascending order of energy. ### Step 1: Final Conclusion The correct ascending order of energy of the given lines is D < A < C < B$D < A < C < B$, corresponding to option (4). ### Pattern Recognition Sees: hydrogen spectral lines energy comparison. Trap: Confusing series limits with specific line numbers. Shortcut: Evaluate transition frequencies or wavelength gaps using Rydberg formula equivalents. ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula Delta E = 13.6 Z^2 left(frac1n_1^2 - frac1n_2^2right)text eV$$\Delta E = 13.6 Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\text{ eV}$$ ### Core Logic Step 1: First Balmer line (n_1 = 2, n_2 = 3$n_1 = 2, n_2 = 3$): Delta E_1 = x = 13.6 left(frac12^2 - frac13^2right) = 13.6 left(frac14 - frac19right) = 13.6 left(frac536right)$$\Delta E_1 = x = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \left(\frac{5}{36}\right)$$ Step 2: Second Balmer line (n_1 = 2, n_2 = 4$n_1 = 2, n_2 = 4$): Delta E_2 = 13.6 left(frac12^2 - frac14^2right) = 13.6 left(frac14 - frac116right) = 13.6 left(frac316right)$$\Delta E_2 = 13.6 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \left(\frac{3}{16}\right)$$ Step 3: Ratio of energies: fracDelta E_2x = fracfrac316frac536 = frac3 times 3616 times 5 = frac2720 = 1.35$$\frac{\Delta E_2}{x} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{3 \times 36}{16 \times 5} = \frac{27}{20} = 1.35$$ Delta E_2 = 1.35 x$$\Delta E_2 = 1.35 x$$ ### Pattern Recognition Sees: Ratio of hydrogen spectrum spectral line energies. Shortcut: (3/16) / (5/36) = 27/20 = 1.35$(3/16) / (5/36) = 27/20 = 1.35$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom Class 12 Physics: Atoms

### Related Formula psi_n,l,m(r, theta, phi) = R_n,l(r) cdot Y_l,m(theta, phi)$$\psi_{n,l,m}(r, \theta, \phi) = R_{n,l}(r) \cdot Y_{l,m}(\theta, \phi)$$ ### Core Logic Let us evaluate each statement carefully: - **Statement (A)** is **True**: The subsidiary (or azimuthal) quantum number l$l$ dictates the orbital shape (l=0 implies textspherical s$l=0 \implies \text{spherical } s$, l=1 implies textdumbbell p$l=1 \implies \text{dumbbell } p$, l=2 implies textdouble-dumbbell d$l=2 \implies \text{double-dumbbell } d$, etc.).
- **Statement (B)** is **True**:

The diagram clearly displays a boundary surface showing two symmetrical lobes situated along the x-axis, which is the exact depiction of a 2mathrmp_x$2\mathrm{p_x}$ orbital.
- **Statement (C)** is **False**: The +$+$ and -$-$ signs in orbital diagrams indicate the mathematical sign (or phase) of the spatial wave function psi$\psi$ in those regions, not electric charge.
- **Statement (D)** is **False**: The wave function of a 2mathrmp_x$2\mathrm{p_x}$ orbital is zero everywhere in its nodal plane. For a 2mathrmp_x$2\mathrm{p_x}$ orbital, the nodal plane is the **yz-plane** (i.e., at x=0$x=0$), not the xy-plane. ### Step 1: Conclusion Only statements (A) and (B) are correct. ### Pattern Recognition Nodal planes of p-orbitals are perpendicular to the orbital axis: - p_x rightarrow yz$p_x \rightarrow yz$-plane is node (x=0$x=0$) - p_y rightarrow xz$p_y \rightarrow xz$-plane is node (y=0$y=0$) - p_z rightarrow xy$p_z \rightarrow xy$-plane is node (z=0$z=0$) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom