Question

The energy level diagram of an element is given blow. Identify by doing necessary calculation which transition corresponds to the emission of a spectral line of wavelength $102.7nm$

Answer 1

From the given energy level diagram of an element for element A

$E_1=-1.5ev$

$E_2=0.85ev$

Energy of emitted photon is

$E=\frac{hc}{\lambda }$

$\lambda =\frac{hc}{E}$

$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{0.65\times 1.6\times {10}^{-19}}$

$=19.038\times {10}^{-7}$

$=1903.8nm$

For element B

$E_1=-3.4ev$

$E_2=-0.85ev$

$E_2-E_1=2.55ev$

$\lambda =\frac{hc}{E}$

$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2.55\times 1.6\times {10}^{-19}}$

$=4.8529\times {10}^{-7}$

$=485.2nm$

For element C

$E_1=-3.4ev$

$E_2=-1.5ev$

$E=E_2-E_1=1.9ev$

$\lambda =\frac{hc}{E}$

$\lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.9\times 1.6\times {10}^{-19}}$

$\lambda =6.5131\times {10}^{-7}$

$=651nm$

For element D

$E_1=-13.6ev$

$E_2=-1.5ev$

$E=E_2-E_1=12.1ev$

$\lambda =\frac{hc}{E}$

$\lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{12.1\times 1.6\times {10}^{-19}}$

$\lambda =1.0227\times {10}^{-7}$

$=102.27nm$

Hence, the transition from -1.5 ev to -13.6ev for element D corresponds to the emission of a spectral line of wavelength 102.7 nm