The value of Rydberg constant (R_H) is 2.18×10^–18 J. The velocity of electron having mass 9.1×10^–31 kg in Bohr's first orbit of hydrogen atom = ……… ×10⁵ ms^–1 (nearest integer)

To find the velocity of the electron in the first Bohr orbit of a hydrogen atom, we can use the expression for the energy of the electron and its relation to velocity.

The Rydberg constant in joules represents the ionization energy of a hydrogen atom in its ground state, which corresponds to the magnitude of the energy of the electron in the first orbit (n = 1).
The energy of an electron in the n-th orbit of a hydrogen-like atom is given by:
$E_n=-\frac{R_H}{n^2}$
For the first orbit of hydrogen (n = 1):
$E_1=-2.18\times {10}^{-18}\text{J}$

According to the Bohr model, the kinetic energy (K.E.) of the electron in an orbit is equal to the negative of its total energy (E):
$\text{K.E.}=-E$
Therefore, the kinetic energy of the electron in the first orbit is:
$\text{K.E.}=2.18\times {10}^{-18}\text{J}$

The formula for kinetic energy is:
$\text{K.E.}=\frac{1}{2}m{v}^2$
where:
m is the mass of the electron = $9.1\times {10}^{-31}\text{kg}$
v is the velocity of the electron.

Substituting the values into the kinetic energy equation:
$2.18\times {10}^{-18}=\frac{1}{2}\times (9.1\times {10}^{-31})\times v^2$
Rearranging the equation to solve for v²:
$v^2=\frac{2\times 2.18\times {10}^{-18}}{9.1\times {10}^{-31}}$
$v^2=\frac{4.36}{9.1}\times {10}^{13}$
$v^2\approx 0.47912\times {10}^{13}=4.7912\times {10}^{12}{\text{m}}^2{\text{s}}^{-2}$

Taking the square root to find v:
$v=\sqrt{4.7912\times {10}^{12}}\approx 2.1889\times {10}^6{\text{ms}}^{-1}$

To express this in the form $\dots \dots \times {10}^5{\text{ms}}^{-1}$:
$v\approx 21.889\times {10}^5{\text{ms}}^{-1}$

Rounding 21.889 to the nearest integer gives 22.