What is the relationship between the wavelength of a wave and its energy?
2 answers:
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Answer:
a)The approximate radius of the nucleus of this atom is 4.656 fermi.
b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Explanation:
= Constant for all nuclei
r = Radius of the nucleus
A = Number of nucleons
a) Given atomic number of an element = 25
Atomic mass or nucleon number = 52
The approximate radius of the nucleus of this atom is 4.656 fermi.
b)
k= = Coulombs constant
= charges kept at distance 'a' from each other
F = electrostatic force between charges
Force of repulsion between two protons on opposite sides of the diameter
The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
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The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
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● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
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● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
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● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
