The ground state orbit in the Bohr atom model of the hydrogen atom has a radius

Question

The ground state orbit in the Bohr atom model of the hydrogen atom has a radius of 5.2912 x 10^-11 m (about O.53A). Although the Bohr model is no longer considered an adequate model of the atom, the Bohr radius often shows up as a useful constant in quantum mechanics. (d) Determine the circumference of the orbit and compare it to the wavelength of the electron. (e) What might you expect the radius to be for the next higher energy orbit (base this on your previous answer)? (f) Determine the wavelength of a 90 mph fastball (a baseball has as a mass of 145 grams). (g) Explain. from your results, why we don't worry about the wave properties of a baseball, even though it has a wavelength when it is moving.

Explanation / Answer

(d)

Radius of the orbit, r = 5.29*10-11 m

So, circumference of the orbit = 2*pi*r = 2*3.14*5.29*10-11 = 33.22*10-11 m

Velocity of electron in this orbit, v = 2.19*106 m/s

Mass of electron, m = 9.1*10-31 kg

Planck's constant, h = 6.626*10-34 m2 kg/s

So, wavelength of electron, l = h/(mv) = 0.332*10-9 m = 33.2*10-11 m

Thus we see that the circumference and wavelength of electron have essentially the same magnitude

(e)

Radius of 'n'th orbit in Bohr's model is given by :

r = 5.29*10-11*n2 m

For the next orbit, n = 2, so

r = 21.16*10-11 m, that is 4 times the Bohr's radius

(f)

Mass of ball, m = 145g = 0.145 kg

Speed of ball, v = 90 mph = 40.2335 m/s

So, Wavelength ( called De-Broglie wavelength) of ball = h/(mv) = 1.135*10-34 m

(g)

As we can see from the previous answer, even though the ball possess some wavelength at this velocity, the magnitude of this wavelength is so so small as to even take it into consideration. Only when the ball starts moving with a speed comparable to that of light, its wavelength becomes appreciable. But for normal cases, it is extremely negligible

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