The answer to the b) part of this problem is \"the kineticenergy is too large to

Question

The answer to the b) part of this problem is "the kineticenergy is too large to expect that the electron could be confinedto a region the size of the nucleus". Could you please offer additional comment on this answer as Ido not understand this statement. Thank you. The answer to the b) part of this problem is "the kineticenergy is too large to expect that the electron could be confinedto a region the size of the nucleus". Could you please offer additional comment on this answer as Ido not understand this statement. Thank you.

Explanation / Answer

For a nucleus of radius 'a', the diameter is 2a. For an electron to be confined to this diameter, the maximumpermissible uncertainty in position is : x < 2a => 1/x > 1/2a According to Heisenberg, x.p >= h/4 So, p >= (h/4*2a) So, p >= h/8a So, However, p = h/ Here, = nL, if L is the circumference of the nucleus. So, maximum value of h/ is h/L Hence, p

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