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 $H{e}^{+}$ion is $\frac{9x}{5}$
(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 $4d_{x^2-y^2}$ 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
$H{e}^{+},n_1=2,n_2=3$
$\frac{1}{\lambda }=R_H(2{)}^2\left[\frac{1}{(2{)}^2}-\frac{1}{(3{)}^2}\right]$
$\frac{1}{\lambda }=\frac{1}{x}4\left[\frac{1}{4}-\frac{1}{9}\right]$
$\frac{1}{\lambda }=\frac{1}{x}\times \frac{5}{9}$
$\lambda =\frac{9x}{5}$
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 $=n-l-1$
$=4-2-1$
$=1$
4. Spin quantum number cannot be obtained from Schrodinger wave equation.