The ionisation enthalpy of hydrogen atom is $1.312 \times 10^{6}$ J $m o l^{-}$ . The energy required to excite the electron in the atom from n=1 to n=2 is:

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Solution

The correct option is A
$E = (I E)_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = 1.312 \times 10^{6} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = 9.84 \times 10^{5} J m o l^{-}$

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Similar questions

Q. The ionization enthalpy of hydrogen atom is

1.312 \times 10^{6} J m o l^{- 1}

. The energy required to excite the atom from n = 1 to n = 2 is:

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1.312 \times 10^{6} J (m o l)]

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The energy of an electron in the ground state of a $H$ -atom is $- 2.15 \times 10^{- 18} J$ . The Ionisation energy of the atomic hydrogen in terms of $10^{5} J m o l^{- 1}$ is:

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The ionisation enthalpy of hydrogen atom is 1.312×106 J mol− . The energy required to excite the electron in the atom from n=1 to n=2 is:

The correct option is A E=(IE)H(1n21−1n22)=1.312×106(112−122)=9.84×105Jmol−

The ionisation enthalpy of hydrogen atom is $1.312 \times 10^{6}$ J $m o l^{-}$ . The energy required to excite the electron in the atom from n=1 to n=2 is:

The correct option is A
$E = (I E)_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = 1.312 \times 10^{6} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = 9.84 \times 10^{5} J m o l^{-}$