An ion has a negative charge
Answer:
The edge length is 0.4036 nm
Solution:
As per the question:
Density of Ag, 
Density of Pd, 
Atomic weight of Ag, A = 107.87 g/mol
Atomic weight of Pd, A' = 106.4 g/mol
Now,
The average density, 
where
= Volume of crystal lattice
a = edge length
n = 4 = no. of atoms in FCC
Therefore,
Therefore, the length of the unit cell is given as:
(1)
Average atomic weight is given as:
where
= 79 %
= 107
= 21%
= 106
Therefore,
In the similar way, average density is given as:
Therefore, edge length is given by eqn (1) as:
Here are a list of items I found.
some brushes
a armature
a permanent magnet
some slip rings
Answer:
0.5 s
Explanation:
From the question given above, the following data were obtained:
Number of circle (n) = 2
Time (t) = 1 s
Period =?
Period of a wave is simply defined as the time taken to make one complete oscillation. Mathematically, it can be expressed as:
T = t / n
Whereb
T => is the period
t => is the space time
n => is the number of circle or oscillation.
With the above formula, we can obtain the period of the wave as follow:
Number of circle (n) = 2
Time (t) = 1 s
Period =?
T = t / n
T = 1 / 2
T = 0.5 s
Thus, the period of the wave is 0.5 s
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
