Question

A monochromatic radiation of specific wavelength is incident on $H$ atoms in ground state. $H$ atoms absorb energy and subsequently emit radiations of six different wavelength. Find the wavelength of incident radiation.

• A. $587.6\stackrel{\circ }{\text{A}}$
• B. $648.2\stackrel{\circ }{\text{A}}$
• C. $972.5\stackrel{\circ }{\text{A}}$
• D. $1050.68\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is C $972.5\stackrel{\circ }{\text{A}}$

We know, number of spectral lines = $\frac{n(n-1)}{2}$
where, n is the orbit number.
Six spectral lines are emitted,
$\frac{n(n-1)}{2}$ = 6
$n^2-n-12=0$.
On solving, n = 4, -3 but n = -3 is not possible
Energy in an orbit $E_n$ = $-13.6\frac{Z^2}{n^2}$
where, Z = Atomic number
Difference in Energy from ground state to excited state = $E_{n=4}-E_{n=0}$
$\Delta E=-13.6\frac{1}{4^2}-(-13.6\frac{1}{1^2}$) =
$\Delta E$ = 13.6 - 0.85 eV = 12.75 eV
According to Planck's quantum theory E = $\frac{12400}{\lambda }$ eV
where, $\lambda$ = wavelength of radiation in $\stackrel{\circ }{\text{A}}$
12.75 = $\frac{12400}{\lambda }$
$\lambda$ = $\frac{12400}{12.75}\stackrel{\circ }{\text{A}}$ = 972.5 $\stackrel{\circ }{\text{A}}$