Question

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 $He ^{+}$ion is $\frac{9 x}{5}$
(2) When an electron jumps from $4\text{ th}$ 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 atom

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (D)

$\frac{1}{\lambda}=R_{H} Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
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
$He ^{+}, n_{1}=2, n_{2}=3$
$\frac{1}{\lambda}=R_{ H }(2)^{2}\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]$
$\frac{1}{\lambda}=\frac{1}{x} 4\left[\frac{1}{4}-\frac{1}{9}\right]$
$\frac{1}{\lambda}=\frac{1}{x} \times \frac{5}{9}$
$\lambda=\frac{9 x}{5}$
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 $=n-l-1$
$=4-2-1 $
$=1$
4. Spin quantum number cannot be obtained from Schrodinger wave equation.