The electron energy of hydrogen atom in ...

The electron energy of hydrogen atom in the ground state works out to be ` - 2.18 xx 10^(-18)` J per atom. Calculate what will happen to the position of the electron in this atom if an energy of `1.938 xx 10^(-18)`J is supplied to the each hydrogen atom.

Text Solution

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Energy of H atom in the ground state `= -2.18 xx 10^(-18) J "atom"^(-1)`
Energy added =` 1.938 xx 10^(-18) J "atom"^(-1)`
Energy of electron in the excited state `= (-2.18 + 1.938) xx 10^(-18) J "atom"^(-1) = -0.242 xx 10^(-18) J "atom"^(-1)`
` therefore - 0.242 xx 10^(-18) J "atom"^(-1) = (-2.18 xx 10^(-18) J "atom"^(-1))/(n^2)`
`n^2 = (-2.18 xx 10^(-18) J "atom"^(-1) )/(-0.242 xx 10^(-18) J "atom"^(-1) ) = 9 , n = 3`
Hence electron will get excited to third shell.

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zero

13.6 eV

`-13.6 eV`

`-13.6 J`

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1505 kj\mole

1310 kj\mole

1608 kj\mole

None

The radius of the orbit of an electron in the ground state of hydrogen atom is a_0 . Then

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the radius of orbit of an electron in the third energy state of `He^(+)` ion (Z=2) is `3a_0`

the radius orbit of an electron in the ground state of `Li^(2+)` ion (Z=3) is `(a_0)/(3)`

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