An excited He^{+} ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its intial excited state is (for photon of wavelength \lambda, energy E=\frac{124eV}{\lambda}): 

  • Option 1)

    n=5

  • Option 2)

    n=6

  • Option 3)

    n=7

  • Option 4)

    n=4

 

Answers (1)

 

Energy emitted due to transition of electron -

\Delta E= Rhcz^{2}\left ( \frac{1}{n_{f}\, ^{2}}-\frac{1}{n_{i}\, ^{2}} \right )

\frac{1}{\lambda }= Rz^{2}\left ( \frac{-1}{n_{i}\, ^{2}}+\frac{1}{n_{f}\, ^{2}} \right )

- wherein

R= R hydberg\: constant

n_{i}= initial state \\n_{f}= final \: state

 

 

Total energy in initial excited state = \frac{1240}{108.5}+\frac{1240}{30.4}

                                                       =52.22eV

                      52.22eV=13.6\times4\left[1-\frac{1}{n^{2}} \right ]

                      0.96=1-\frac{1}{n^{2}}

                     \Rightarrow \frac{1}{n^{2}}=0.04

                    \Rightarrow n^{2}=\frac{1}{0.04}=25

                    \Rightarrow n=5


Option 1)

n=5

Option 2)

n=6

Option 3)

n=7

Option 4)

n=4

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