Radiation coming from transitions n=2 to n=1 of hydrogen atoms fall on He^{+} ions in n=1 and n=2 states. The possible transition of helium ions as they adsorb energy from the radiation is :

  • Option 1)

    n=2\rightarrow n=5

  • Option 2)

    n=1\rightarrow n=4

  • Option 3)

    n=2\rightarrow n=4

     

  • Option 4)

    n=2\rightarrow n=3

 

Answers (1)

For H- atom

Z=1

E_{1}=13.6\times1(\frac{1}{1^{2}}-\frac{1}{2^{2}})=13.6\times\frac{3}{4}

\\For \; He^{+}\: ions\\Z=2

and n_{1}=1\; or\; n_{1}=2

E_{2}=13.6\times(2^{2})\left (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

equate E_{1}=E_{2}

\\13.6\times\frac{3}{4}=13.6\times4\left (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )\\\Rightarrow\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} =\frac{3}{16}

for \; n_{1}=2

\frac{1}{n{_{2}}^{2}}=\frac{1}{4}-\frac{3}{16}=\frac{1}{16}

n_{2}=4

so possible transition will be

n=2\rightarrow n=4


Option 1)

n=2\rightarrow n=5

Option 2)

n=1\rightarrow n=4

Option 3)

n=2\rightarrow n=4

 

Option 4)

n=2\rightarrow n=3

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