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:
Answer:
λ = 1.4 × 10^(-7) m
Explanation:
We are given;
distance of eye piece from the source;D = 1.5 m
distance between the virtual sources;d = 7.5 × 10^(-4) m
To find the wavelength, we will use the formula for fringe width;
X = λD/d
Where X is fringe width, λ is wavelength, while d and D remain as before.
Now, fringe width = eye-piece distance moved transversely/number of fringes
Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m
Thus,
Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m
Thus;
1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))
λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5
λ = 1.4 × 10^(-7) m
I believe the correct answer from the choices listed above is the second option. The scientific notation of the measurement 0.00000000062 kg would be <span>6.2 x 10^-10 kg. Scientific notation is used to express too large and too small values of numbers. Hope this helps. Have a nice day.</span>
