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Kinetic energy or velocity of the electron in a Bohr orbit

This type of question, asked 3 times: 2021 Paper 20, 5 Apr 2024, 6 Apr 2024. All 3 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 20

    53
    The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to h2xma02\frac{h^{2}}{xma_{0}^{2}}. The value of 10x10x is. (a0\left( a_{0} \right. is radius of Bohr's orbit) (Nearest integer) [Given: π=3.14\pi= 3.14 ]

  2. Question 2

    5 Apr 2024

    51
    The value of Rydberg constant (RH)\left( R_{H} \right) is 2.18×10−18 J2.18 \times10^{- 18}\text{ }J. The velocity of electron having mass 9.1×10−31 kg9.1 \times10^{- 31}\text{ }kg in Bohr's first orbit of hydrogen atom =………×105 ms−1= \ldots\ldots\ldots \times10^{5}{\text{ }ms}^{- 1} (nearest integer)

  3. Question 3

    6 Apr 2024

    149
    For hydrogen atom, energy of an electron in first excited state is −3.4eV,K.E- 3.4eV,K.E. of the same electron of hydrogen atom is xeVxeV. Value of xx is ____ ×10−1eV\times10^{- 1}eV. (Nearest integer)