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The radius of hydrogen atom in its groun...

The radius of hydrogen atom in its ground state is `5.3 xx 10^(-11)m`. After collision with an electron, it is found to have a radius of `21.2 xx 10^(-11)m`. What is the principal quantum number of the final state of the atom?

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In the ground state of hydrogen atom orbital radius is 5.3 xx 10^(-11) m. The atom is excited such that atomic radius becomes 21.2 xx 10^(-11) m . What is the principal quantum number of the excited state of atom ?

In the ground state of hydrogen atom , its Bohr radius is given as 5.3 xx 10^(-11) m. The atom is excited such athat th radius becomes 21.2 xx 10^(-11) m. Find (i) the value of principal quantum number , and (ii) the total energy of the atom in this excited state.

Knowledge Check

  • The radius of hydrogen atom in its ground state is 5.3 xx 10^(-11)m . After collision with an electron it is found to have a radius of 21.2 xx 10^(-11)m . What is the principle quantum number of n of the final state of the atom ?

    A
    `n = 4`
    B
    `n = 2`
    C
    `n = 16`
    D
    `n = 3`

  • The radius of hydrogen atom in its ground state is 5.3 xx 10^-11 m . After collision with an electron it is found to have a radius of 21.2 xx 10^-11 m . The principal quantum number of the final state of the atom is.

    A
    2
    B
    3
    C
    4
    D
    5

  • The radius of hydrogen atom, in its ground state, is of the order of

    A
    `10^(-8)` cm
    B
    `10^(-6)` cm
    C
    `10^(-5)` cm
    D
    `10^(-4)` cm

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    In the ground state of hydrogen atom, its Bohr radius is 5.3xx10^(-11)m . The atom is excited such that the radius becomes 21.2xx10^(-11)m . Find the value of principal quantum number and total energy of the atom in excited state.

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    The radius of the hydrogen atom in its ground state is 5.3xx10^(-11) m. The principal quantum number of the final state of the atom is

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