The uncertainty in an electron\'s position is 0.14nm . Part A What is the minimu
ID: 2262246 • Letter: T
Question
The uncertainty in an electron's position is 0.14nm .Part A What is the minimum uncertainty ?p in its momentum? Express your answer using two significant figures. ?pmin = kg?m/s
Part B What is the kinetic energy of an electron whose momentum is equal to this uncertainty (?p=p)? Express your answer using two significant figures. K = eV
The uncertainty in an electron's position is 0.14nm .
Part A What is the minimum uncertainty ?p in its momentum? Express your answer using two significant figures. ?pmin = kg?m/s
Part B What is the kinetic energy of an electron whose momentum is equal to this uncertainty (?p=p)? Express your answer using two significant figures. K = eV
Explanation / Answer
1. as we know dat from heisenbergs uncertainity principle we ve
?x * ?P = h/2pie
given uncertainty in an electron's position is 0.14nm = 0.14 * 10^-9 m
hence putting the value we ve
0.14 * 10^-9 m * ?P = h/2pie
?P = 6.6 * 10^-34/ 2 pie * 0.14 * 10^-9 = 7.503 * 10^-25 kgm/sec
for the second part we ve
Kinetic energy of the proton K = p2 /2m
= (7.503 * 10^-25)^2/ (2* 1.67*10^-27) = 1.685 * 10^-22 Joule
K = ( 1.685*10-23 J ) ( 1eV / 1.6*10-19 J ) = 1.05342*10-3 eV
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