Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Select the correct statement(s). (1) If the shortest wavelength of H-atom in Lyman series is x, then the longest wavelength in Balmer series of\[H{{e}^{+}}\]ion is\[\frac{9x}{5}\] (2) When an electron jumps from 4th excited state to ground state, maximum number of photons that may be emitted are 4 (3) Number of nodal planes and radial nodes in\[4{{d}_{{{x}^{2}}-{{y}^{2}}}}\]orbital are 2 and 1 respectively (4) All four quantum numbers are obtained from Schrodinger wave equation when applied on atomA) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: D
Solution :
\[\frac{1}{\lambda }={{R}_{H}}{{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] In Lyman series, for shortest wavelength, \[{{n}_{1}}=1,\,{{n}_{2}}=\infty \] \[\frac{1}{x}={{R}_{H}}\times {{(1)}^{2}}\left[ \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(\infty )}^{2}}} \right]\] \[\therefore \] \[{{R}_{H}}=\frac{1}{x}\] In Balmer series, for longest wavelength of \[H{{e}^{+}},{{n}_{1}}=2,{{n}_{2}}=3\] \[\frac{1}{\lambda }={{R}_{H}}{{(2)}^{2}}\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]\] \[\frac{1}{\lambda }=\frac{1}{x}4\left[ \frac{1}{4}-\frac{1}{9} \right]\] \[\frac{1}{\lambda }=\frac{1}{x}\times \frac{5}{9}\] \[\lambda =\frac{9x}{5}\] 2. Number of spectral lines\[=\frac{n(n-1)}{2}\] \[=\frac{4(4-1)}{2}=6\] 3. Nodal planes\[=l=2\](\[\because \]For d orbital,\[l=0\]) Number of radial nodes\[=n-1-1\] \[=4-2-1\] \[=1\] 4. Spin quantum number cannot be obtained from Schrodinger wave equation. Thus, statements (1) and (3) are correct.
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