Question

The energy levels of a hypothetical one electron atom system are given by $E_n=-\frac{16}{n^2}eV$, where $n=1,2,3,....$ The wavelength of emitted photon corresponding to transition from first excited level to ground level is about.

• A. $1035\stackrel{\circ }{A}$
• B. $1220\stackrel{\circ }{A}$
• C. $3650\stackrel{\circ }{A}$
• D. $690\stackrel{\circ }{A}$

Answer 1

The correct option is A $1035\stackrel{\circ }{A}$

Given,
$E_n=\frac{-16}{n^2}eV$
According to the equation,$E_1=-16eV$
$E_2=-\frac{16}{(2{)}^2}eV$
$=-\frac{16}{4}eV$$=-4eV$
We know that,
$\lambda =\frac{12420}{△E}$
$\lambda =\frac{12420}{|E_2-E_1|}$
$\lambda =\frac{12420}{|-4-(-16)|}$
$\lambda =\frac{12420}{12}$
$\lambda =1035\stackrel{\circ }{A}$.