Which of the following transitions in hydrogen atoms emit photons of highest frequency ?

  • Option 1)

    n=1 \; to\; n=2

  • Option 2)

    n=2 \; to\; n=6

  • Option 3)

    n=6 \; to\; n=2

  • Option 4)

    n=2 \; to\; n=1

 

Answers (1)

As we learnt in

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

 

 

;

h\upsilon _{2\rightarrow 1}= -13.6\left ( \frac{1}{2^{2}}- \frac{1}{1^{2}}\right )eV

                = +13.6\times \frac{3}{4}eV= 10.2\: eV

Emission is n=2 \rightarrow n= 1i.e. higher n to lower n Transition from lower to higher levels are absorption lines

-13.6\left ( \frac{1}{6^{2}}- \frac{1}{2^{2}}\right )= +13.6\times \frac{2}{9}

This is < E_{n=2}\rightarrow E_{n=1}

Correct option is 4.

 


Option 1)

n=1 \; to\; n=2

This is an incorrect option.

Option 2)

n=2 \; to\; n=6

This is an incorrect option.

Option 3)

n=6 \; to\; n=2

This is an incorrect option.

Option 4)

n=2 \; to\; n=1

This is the correct option.

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