Find out the energy of $H$ -atom in first excitation state. The value of permittivity factor $4 π ε_{0} = 1.11264 \times 10^{- 10} C^{2} N^{- 1} m^{- 2}$ .

- 1.3 \times 10^{- 19} j o u l e

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- 32 \times 10^{- 19} j o u l e

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- 5.443 \times 10^{- 19} j o u l e

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- 20 \times 10^{- 19} j o u l e

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Solution

The correct option is C $- 5.443 \times 10^{- 19} j o u l e$
In $M . K . S .$ system,
$E_{n} = - \frac{2 π^{2} Z^{2} m e^{4}}{(4 π e_{0})^{2} n^{2} h^{2}}$ $∵ n = 2$
$= - \frac{2 \times (3.14)^{2} \times (1)^{2} \times 9.108 \times 10^{- 31} \times (1.602 \times 10^{- 19})^{4}}{(1.11264 \times 10^{- 10})^{2} \times (2)^{2} \times (6.626 \times 10^{- 34})^{2}}$
$= - 5.443 \times 10^{- 10} j o u l e$

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Find out the energy of H-atom in first excitation state. The value of permittivity factor 4πε0=1.11264×10−10C2N−1m−2.

The correct option is C −5.443×10−19jouleIn M.K.S. system,En=−2π2Z2me4(4πe0)2n2h2 ∵n=2=−2×(3.14)2×(1)2×9.108×10−31×(1.602×10−19)4(1.11264×10−10)2×(2)2×(6.626×10−34)2=−5.443×10−10joule

Find out the energy of $H$ -atom in first excitation state. The value of permittivity factor $4 π ε_{0} = 1.11264 \times 10^{- 10} C^{2} N^{- 1} m^{- 2}$ .