Question

The energy levels of a hypothetical one electron atom is given by $E_n=\frac{-18.0}{n^2}eV$ where
n = 1,2,3,.....
Find the wavelength which will not be emitted when these atoms in the ground state are bombarded by electrons that have been accelerated through a potential difference of 16.2 V.

• A. $4950\stackrel{\circ }{A}$
• B. $773\stackrel{\circ }{A}$
• C. $917\stackrel{\circ }{A}$
• D. $733\stackrel{\circ }{A}$

Answer 1

The correct option is D $733\stackrel{\circ }{A}$

$E_4-E_1=(-1.125)-(-18.0)$
$=16.875eV>16.2eV$
$E_3-E_1=(-2.0)-(-18.0)=16.0eV<16.2eV$
Hence, the electron can excite this atom from n = 1 to n = 3
${\lambda }_{41}=\frac{12375}{E_4-E_1}=\frac{12375}{16.875}=733\stackrel{\circ }{A}$ will not be emitted.