Q17 / 155 · WBJEE 2020 · Chemistry

The radius of the first Bohr orbit of a hydrogen atom is $0.53 \times 10^{-8} \mathrm{~cm}$. The velocity of the electron in the first Bohr orbit is

  • Correct answer
    A
    $2.188 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}$
  • B
    $4.376 \times 10^{8} \mathrm{~cm} \mathrm{~s}^{-1}$
  • C
    $1.094 \times 10^{8} \mathrm{~cm} \mathrm{~s}^{-1}$
  • D
    $2.188 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}$

Solution

Hint: $r=0.53 \times 10^{-9} \mathrm{~cm}=0.53 \times 10^{-9} n^{2}$ $\therefore \mathrm{n}=1$ $\mathrm{~V}=2.18 \times 10^{9} \mathrm{cm} / \mathrm{s}$

Hence, the correct answer is option: 'A'.

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