Chapter 4: Q57E (page 138)

In the Bohr model of the hydrogen atom, the electron can have only certain velocities. Obtain a formula for the allowed velocities, then obtain a numerical value for the highest speed possible.

Short Answer

Expert verified

The equation for the allowed velocity of the electron is $\frac{q^{2}}{4 π ε_{0} / n}$ and the numeric value of the maximum allowed velocity is_{role="math" localid="1659758061348" $2.18 \times 10^{6} m / s$}.

Step by step solution

Bohr Model.

The expression for the velocity for the electron in the Bohr model of hydrogen is given by,

$v^{2} = \frac{q^{2}}{4 {πε}_{0} m r} \frac{1}{r}$

The expression for the radius of the hydrogen atom of the Bohr model is given by,

$r = \frac{(4 {πε}_{0}) d^{2} n^{2}}{m q^{2}}$

Use the expression of velocity and radius for calculation.

The expression for the velocity of the electron is evaluated as,

$v^{2} = \frac{q^{2}}{4 π ε_{0} m r} \frac{1}{(\frac{(4 π ε_{0}) d^{2} n^{2})}{m q^{2}})} = \frac{q^{2}}{4 π ε_{0} / n}$

The velocity of the electron is calculated as,

role="math" localid="1659758442389" $v = \frac{q^{2}}{4 π ε_{0} h n} = \frac{{(1.6 \times 10^{- 9} C)}^{2}}{4 π (8.85 \times 10^{- 12} C^{2} / N . m^{2}) (1.0546 \times 10^{- 34} J . s) (1)} = 2.18 \times 10^{6} m / s$

Therefore, the equation for the allowed velocity of the electron is $\frac{q^{2}}{4 {πε}_{0} / n}$ and the numeric value of the maximum allowed velocity is_{$2.18 \times 10^{6} m / s$}.