What is the electric force on a proton 2.5 fm from the surface of the nucleus? hint: treat the spherical nucleus as a point char
1 answer:
In this case, you need the formula below where:
F = force
k = coulombs constant 8.99 x10^{9} N.m^{2} . C^{-2}
q1 = electric charge 1
q2 = electric charge 2
r = the distance between the charges
pls note: make sure your units are correct (in meters etc, not fm (<em>femto-meters</em>)).
Curiously, this question doesn't tell you what atom you are next to the nucleus of. Different numbers of protons in the nucleus of the atom will make for vastly different forces in your answer...
F = force
k = coulombs constant 8.99 x10^{9} N.m^{2} . C^{-2}
q1 = electric charge 1
q2 = electric charge 2
r = the distance between the charges
pls note: make sure your units are correct (in meters etc, not fm (<em>femto-meters</em>)).
Curiously, this question doesn't tell you what atom you are next to the nucleus of. Different numbers of protons in the nucleus of the atom will make for vastly different forces in your answer...
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Answer:
Explanation:
The formula for hydrogen atomic spectrum is as follows
energy of photon due to transition from higher orbit n₂ to n₁
For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc
energy of first line
10.2 eV
wavelength of photon = 12375 / 10.2 = 1213.2 A
energy of 2 nd line
= 12.08 eV
wavelength of photon = 12375 / 12.08 = 1024.4 A
energy of third line
12.75 e V
wavelength of photon = 12375 / 12.75 = 970.6 A
energy of fourth line
= 13.056 eV
wavelength of photon = 12375 / 13.05 = 948.3 A
energy of fifth line
13.22 eV
wavelength of photon = 12375 / 13.22 = 936.1 A