Question

The ionisation energy of a H-like Bohr atom is 4 Rydbergs, then the wavelength of radiation emitted when the $e^{-}$ jumps from the first excited state to the ground state is: [1Rydberg = $2.18\times {10}^{-18}J]$

• A. $303.89\stackrel{o}{A}$
• B. $304.01\stackrel{o}{A}$
• C. $323.43\stackrel{o}{A}$
• D. $313.34\stackrel{o}{A}$

Answer 1

Answer: (A)

$303.89\stackrel{o}{A}$

The ionization energy is given by
$IE=E_{\infty }-E_1=\Delta E$
$13.6=0-E_1$
$E_1=-4R\left(energyingroundstate\right)$ where $R=Rydberg=2.18\times {10}^{-18}J$
Energy in $n^{th}$ level is given by
$E_n=\frac{E_1}{n^2}$
Energy in first excited state ($n=2$) is given by
$E_2=\frac{E_1}{2^2}=\frac{-4R}{4}=-R$
$\Delta E=E_2-E_1=-R-(-4R)=3R$
$\frac{hc}{\lambda }=3R$
$\Rightarrow \lambda =\frac{hc}{3R}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3\times 2.18\times {10}^{-18}}=303.89A^o$