Question

A. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit?

Answer 1

Answer:

a

  $n = 23$

b

  $v = 87377.95 \ m/s$

Explanation:

From the question we are told that

   The diameter is $d = 61\ nm = 61 *10^{-9} \ m$

   

Generally the radius electron orbit  is mathematically represented as

      $r = \frac{61 *10^{-9}}{2}$

=>   $r = 3.05*10^{-8} \ m$

This radius can also be represented mathematically  as

      $r = n^2 * a_o$

Here n is the quantum number and $a_o$ is  the Bohr radius with a value

    $a_o = 0.0529 *10^{-9} \ m$

So

   $n = \sqrt{\frac{3.05*10^{-8}}{ 0.059*10^{-9}} }$

=>   $n = 23$

Generally the angular momentum of the electron is mathematically represented as

          $L = m * v * r = \frac{n * h }{2 \pi}$

Here  h is the Planck constant and the value is  $h = 6.626*10^{-34} J \cdot s$

          m is the mass of the electron with values $m = 9.1*10^{-31} \ kg$

         So

               $v = \frac{23 * 6.626*10^{-34} }{2\pi * 9.1 *10^{-31} * 3.05*10^{-8} }$

                $v = 87377.95 \ m/s$